{"id":596,"date":"2025-10-09T15:24:43","date_gmt":"2025-10-09T15:24:43","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=596"},"modified":"2025-10-23T13:42:24","modified_gmt":"2025-10-23T13:42:24","slug":"limits-and-continuity-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/limits-and-continuity-learn-it-1\/","title":{"raw":"Limits and Continuity: Learn It 1","rendered":"Limits and Continuity: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Calculate the limit of a function of two variables<\/li>\r\n \t<li>Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach<\/li>\r\n \t<li>Determine when a function is continuous and check the conditions for continuity<\/li>\r\n \t<li>Find limits and check continuity for functions of three or more variables<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Limits<\/h2>\r\n<p class=\"whitespace-normal break-words\">We have now examined functions of more than one variable and seen how to graph them. In this section, we see how to take the limit of a function of more than one variable, and what it means for a function of more than one variable to be continuous at a point in its domain. It turns out these concepts have aspects that just don't occur with functions of one variable.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Limit of a Function of One Variable<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Let [latex]f(x)[\/latex] be defined for all [latex]x\\neq a[\/latex] in an open interval containing [latex]a[\/latex]. Let [latex]L[\/latex] be a real number. Then<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\text{lim}}f(x)=L[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">if for every [latex]\\epsilon&gt;0[\/latex], there exists a [latex]\\delta&gt;0[\/latex], such that if [latex]0&lt;|x-a|&lt;\\delta[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex], then<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]|f(x)-L|&lt;\\epsilon[\/latex].<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">Before we can adapt this definition to define a limit of a function of two variables, we first need to extend the idea of an open interval in one variable to an open interval in two variables.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>[latex]\\delta[\/latex] disk<\/h3>\r\n<p class=\"whitespace-normal break-words\">A <strong>[latex]\\delta[\/latex] disk<\/strong> centered at point [latex](a,\\ b)[\/latex] is an open disk of radius [latex]\\delta[\/latex] centered at point [latex](a,\\ b)[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{(x,y)\\in\\mathbb{R}^{2},|,(x-a)^{2}+(y-b)^{2}&lt;\\delta^{2}}[\/latex]<\/p>\r\n\r\n<\/section>[caption id=\"attachment_970\" align=\"aligncenter\" width=\"423\"]<img class=\"size-full wp-image-970\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28223402\/4-2-1.jpeg\" alt=\"On the xy plane, the point (2, 1) is shown, which is the center of a circle of radius \u03b4.\" width=\"423\" height=\"197\" \/> Figure 1. A [latex]\\delta[\/latex] disk centered around the point [latex](2,1)[\/latex].[\/caption]\r\n<p class=\"whitespace-normal break-words\">The idea of a [latex]\\delta[\/latex] disk appears in the definition of the limit of a function of two variables. If [latex]\\delta[\/latex] is small, then all the points [latex](x,\\ y)[\/latex] in the [latex]\\delta[\/latex] disk are close to [latex](a,\\ b)[\/latex]. This is completely analogous to [latex]x[\/latex] being close to [latex]a[\/latex] in the definition of a limit of a function of one variable. In one dimension, we express this restriction as<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]a-\\delta&lt;x&lt;a+\\delta[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">In more than one dimension, we use a [latex]\\delta[\/latex] disk.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>limit of a function of two variables<\/h3>\r\n<p class=\"whitespace-normal break-words\">Let [latex]f[\/latex] be a function of two variables. The <strong>limit<\/strong> of [latex]f(x,y)[\/latex] as [latex](x,\\ y)[\/latex] approaches [latex](a,\\ b)[\/latex] is [latex]L[\/latex], written:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{(x,y)\\to(a,b)}{\\text{lim}}f(x,y)=L[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">if for each [latex]\\epsilon&gt;0[\/latex] there exists a [latex]\\delta&gt;0[\/latex] such that for all points [latex](x,\\ y)[\/latex] in a [latex]\\delta[\/latex] disk around [latex](a,\\ b)[\/latex], except possibly for [latex](a,\\ b)[\/latex] itself, we have:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]|f(x,y)-L|&lt;\\epsilon[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">In symbols:<\/p>\r\n&nbsp;\r\n<p class=\"whitespace-normal break-words\">For any [latex]\\varepsilon&gt;0[\/latex], there exists [latex]\\delta&gt;0[\/latex] such that [latex]|f(x,y)-L|&lt;\\varepsilon[\/latex] whenever [latex]0&lt;\\sqrt{(x-a)^{2}+(y-b)^{2}}&lt;\\delta[\/latex]<\/p>\r\n\r\n<div><\/div>\r\n<\/section>[caption id=\"attachment_971\" align=\"aligncenter\" width=\"646\"]<img class=\"size-full wp-image-971\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28223617\/4-2-2.jpeg\" alt=\"In xyz space, a function is drawn with point L. This point L is the center of a circle of radius \u0949, with points L \u00b1 \u0949 marked. On the xy plane, there is a point (a, b) drawn with a circle of radius \u03b4 around it. This is denoted the \u03b4-disk. There are dashed lines up from the \u03b4-disk to make a disk on the function, which is called the image of delta disk. Then there are dashed lines from this disk to the circle around the point L, which is called the \u0949-neighborhood of L.\" width=\"646\" height=\"581\" \/> Figure 2.\u00a0The limit of a function involving two variables requires that\u00a0[latex]f(x,y)[\/latex] be within\u00a0[latex]\\varepsilon[\/latex] of\u00a0[latex]L[\/latex]\u00a0whenever [latex](x,y)[\/latex]\u00a0is within [latex]\\delta[\/latex] of\u00a0[latex](a,b)[\/latex]. The smaller the value of\u00a0[latex]\\varepsilon[\/latex],\u00a0the smaller the value of\u00a0[latex]\\delta[\/latex].[\/caption]Proving that a limit exists using the definition of a limit of a function of two variables can be challenging. Instead, we use the following theorem, which gives us shortcuts to finding limits. The formulas in this theorem are extensions of the formulas in the limit laws theorem for single-variable functions.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>limit laws for functions of two variables theorem<\/h3>\r\nLet [latex]\\underset{(x,y)\\to(a,b)}{\\text{lim}}f(x,y)=L[\/latex] and [latex]\\underset{(x,y)\\to(a,b)}{\\text{lim}}g(x,y)=M[\/latex], and let [latex]c[\/latex] be a constant. Then:\r\n\r\n&nbsp;\r\n\r\n<strong>Constant\u00a0Law:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}c=c[\/latex]<\/p>\r\n<strong>Identity Law:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}x} &amp; =\\hfill &amp; {a} \\hfill \\\\ \\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}y} &amp; =\\hfill &amp; {b} \\end{array}[\/latex]<\/p>\r\n<strong>Sum Law:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y)+g\\,(x,\\ y))=L+M[\/latex]<\/p>\r\n<strong>Difference Law:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y)-(g\\,(x,\\ y))=L-M[\/latex]<\/p>\r\n<strong>Constant Multiple<\/strong> <strong>Law:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(cf\\,(x,\\ y))=cL[\/latex]<\/p>\r\n<strong>Product Law:\u00a0<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y)\\,g\\,(x,\\ y))=LM[\/latex]<\/p>\r\n<strong>Quotient Law:\u00a0<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}\\frac{f\\,(x,\\ y)}{g\\,(x,\\ y)}=\\frac{L}{M}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">for [latex]M\\neq{0}[\/latex]<\/p>\r\n<strong>Power Law:\u00a0<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y))^{n}=L^{n}[\/latex]<\/p>\r\nfor any positive integer [latex]n[\/latex].\r\n\r\n<strong>Root Law:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}\\sqrt[n]{f\\,(x,\\ y)}=\\sqrt[n]{L}[\/latex]<\/p>\r\nfor all [latex]L[\/latex] if [latex]n[\/latex] is odd and positive, and for [latex]L\\,\\geq\\,0[\/latex] if [latex]n[\/latex] is even and positive.\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">The proofs of these properties are similar to those for the limits of functions of one variable. We can apply these laws to finding limits of various functions.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Find each of the following limits:\r\n<ol>\r\n \t<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}\\frac{2x+3y}{4x-3y}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1167793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793933114\"]\r\n<ol>\r\n \t<li>First use the sum and difference laws to separate the terms:<\/li>\r\n<\/ol>\r\n[latex]\\hspace{5cm}\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)[\/latex]\r\n<div style=\"text-align: center;\">[latex]\\hspace{5cm}=\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x^{2}\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}2xy\\bigg) +\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y^{2}\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}4x\\bigg) \\\\ \\hspace{5cm}+\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}6\\bigg).[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNext,\u00a0use the constant multiple law on the second, third, fourth, and fifth limits:\r\n<div style=\"text-align: center;\">[latex]\\hspace{5cm}=\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x^{2}\\bigg)-2\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}xy\\bigg)+3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y^{2}\\bigg)-4\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\bigg) \\\\ \\hspace{5cm} +3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}6\\bigg).[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNow, use the power law on the first and third limits, and the product law on the second limit:\r\n<div style=\"text-align: center;\">[latex]\\hspace{5cm} =\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\bigg)^{2}-2\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}xy\\bigg)+3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\bigg)^{2} \\\\ \\hspace{5cm} -4\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\bigg)+3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}6\\bigg).[\/latex]<\/div>\r\n&nbsp;\r\n\r\nLast, use the identity laws on the first six limits and the constant law on the last limit:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill{\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)}&amp; =\\hfill &amp; {(2)^{2}-2(2)(-1)+3(-1)^{2}-4(2)+3(-1)-6} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-6} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n<ol>\r\n \t<li>Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, constant multiple law, and identity law,\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(4x-3y)} &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}4x-\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {4\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\big)-3\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\big)} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {4(2)-3(-1)=11.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nSince the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of the numerator using the difference law, constant multiple law, and identity law:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(2x+3y)} &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}2x+\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\big)+3\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\big)} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2(2)+3(-1)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {1.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTherefore, according to the quotient law we have\r\n<div style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}\\frac{2x+3y}{4x-3y}=\\frac{\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(2x+3y)}{\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(4x-3y)}=\\frac{1}{11}.[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Since we are taking the limit of a function of two variables, the point [latex](a,\\ b)[\/latex] is in [latex]\\mathbb{R}^{2}[\/latex], and it is possible to approach this point from an infinite number of directions. Sometimes when calculating a limit, the answer varies depending on the path taken toward [latex](a,\\ b)[\/latex]. If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of path taken\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Show that neither of the following limits exist:\r\n<ol>\r\n \t<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{2xy}{3x^{2}+y^{2}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1197793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1197793933114\"]\r\n<ol>\r\n \t<li>The domain of the function [latex]f\\,(x,\\ y)=\\frac{2xy}{3x^{2}+y^{2}}[\/latex]\u00a0consists of all points in the [latex]xy[\/latex]-plane\u00a0except for the point [latex](0,\\ 0)[\/latex]\u00a0(Figure 3). To show that the limit does not exist as [latex](x,\\ y)[\/latex] approaches [latex](0,\\ 0)[\/latex],\u00a0we note that it is impossible to satisfy the definition of a limit of a function of two variables because of the fact that the function takes different values along different lines passing through point [latex](0,\\ 0)[\/latex].\u00a0First, consider the line [latex]y=0[\/latex] in the [latex]xy[\/latex]-plane. Substituting\u00a0[latex]y=0[\/latex] into [latex]f\\,(x,\\ y)[\/latex] gives\r\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ 0)=\\frac{2x(0)}{3x^{2}+0^{2}}=0}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nFor any value of [latex]x[\/latex].\u00a0Therefore the value of [latex]f[\/latex]\u00a0remains constant for any point on the [latex]x[\/latex]-axis, and as [latex]y[\/latex] approaches zero, the function remains fixed at zero.\u00a0Next, consider the line [latex]y=x[\/latex]. Substituting [latex]y=x[\/latex] into [latex]f\\,(x,\\ y)[\/latex] gives\r\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ x)=\\frac{2x(x)}{3x^{2}+x^{2}}=\\frac{2x^{2}}{4x^{2}}=\\frac{1}{2}}.[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThis is true for any point on the line [latex]y=x[\/latex]. If we let [latex]x[\/latex] approach zero\u00a0while staying on this line, the value of the function remains fixed at [latex]\\frac{1}{2}[\/latex],\u00a0regardless of how small [latex]x[\/latex] is.\u00a0Choose a value for [latex]\\epsilon[\/latex] that is less than [latex]1\/2-[\/latex]say, [latex]1\/4[\/latex]. Then, no matter how small a [latex]\\delta[\/latex] disk\u00a0we draw around [latex](0,\\ 0)[\/latex], the values of [latex]f\\,(x,\\ y)[\/latex] for points inside that [latex]\\delta[\/latex] disk will include both [latex]0[\/latex] and [latex]\\frac{1}{2}.[\/latex] Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.\r\n\r\n[caption id=\"attachment_973\" align=\"aligncenter\" width=\"451\"]<img class=\"size-full wp-image-973\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28224845\/4-2-3.jpeg\" alt=\"In xyz space, the function f(x, y) = 2xy\/(3x2 + y2) is shown, which is a slightly twisted plane, with values of 0 along the line y = 0 and values of \u00bd along the line y = x.\" width=\"451\" height=\"288\" \/> Figure 3. Graph of the function [latex]\\small{f(x,y)=(2xy)\/(3x^{2}+y^{2})}[\/latex]. Along the line [latex]\\small{y=0}[\/latex], the function is equal to zero; along the line [latex]\\small{y=x}[\/latex], the function is equal to [latex]\\small{\\frac{1}{2}}[\/latex].[\/caption]<\/li>\r\n \t<li>In a similar fashion to 1., we can approach the origin along any straight line passing through the origin. If we try the [latex]x[\/latex]-axis (i.e., [latex]y=0[\/latex]),\u00a0then the function remains fixed at zero. The same is true for the [latex]y[\/latex]-axis.\u00a0Suppose we approach the origin along a straight line of slope [latex]k[\/latex]\u00a0The equation of this line is [latex]y=kx[\/latex]. Then the limit becomes\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}} &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4x(kx)^{2}}{x^{2}+3(kx)^{4}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4k^{2}x^{3}}{x^{2}+3k^{4}x^{4}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4k^{2}x}{1+3k^{4}x^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}4k^{2}x}{\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}(1+3k^{4}x^{2})}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {0}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nRegardless of the value of [latex]k[\/latex].\u00a0It would seem that the limit is equal to zero. What if we chose a curve passing through the origin instead? For example, we can consider the parabola given by the equation [latex]x=y^{2}[\/latex]. Substituting [latex]y^{2}[\/latex] in place of [latex]x[\/latex] in [latex]f\\,(x,\\ y)[\/latex] gives\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}} &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4(y^{2})y^{2}}{(y^{2})^{2}+3y^{4}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4y^{4}}{y^{4}+3y^{4}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}1}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {1.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nBy the same logic\u00a0in 1., it is impossible to find a [latex]\\delta[\/latex] disk\u00a0around the origin that satisfies the definition of the limit for any value of [latex]\\epsilon\\,&lt;\\,1[\/latex]. Therefore, [latex]\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}[\/latex] does not exist.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Calculate the limit of a function of two variables<\/li>\n<li>Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach<\/li>\n<li>Determine when a function is continuous and check the conditions for continuity<\/li>\n<li>Find limits and check continuity for functions of three or more variables<\/li>\n<\/ul>\n<\/section>\n<h2>Limits<\/h2>\n<p class=\"whitespace-normal break-words\">We have now examined functions of more than one variable and seen how to graph them. In this section, we see how to take the limit of a function of more than one variable, and what it means for a function of more than one variable to be continuous at a point in its domain. It turns out these concepts have aspects that just don&#8217;t occur with functions of one variable.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Limit of a Function of One Variable<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Let [latex]f(x)[\/latex] be defined for all [latex]x\\neq a[\/latex] in an open interval containing [latex]a[\/latex]. Let [latex]L[\/latex] be a real number. Then<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{x\\to a}{\\text{lim}}f(x)=L[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">if for every [latex]\\epsilon>0[\/latex], there exists a [latex]\\delta>0[\/latex], such that if [latex]0<|x-a|<\\delta[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex], then<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]|f(x)-L|<\\epsilon[\/latex].<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">Before we can adapt this definition to define a limit of a function of two variables, we first need to extend the idea of an open interval in one variable to an open interval in two variables.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>[latex]\\delta[\/latex] disk<\/h3>\n<p class=\"whitespace-normal break-words\">A <strong>[latex]\\delta[\/latex] disk<\/strong> centered at point [latex](a,\\ b)[\/latex] is an open disk of radius [latex]\\delta[\/latex] centered at point [latex](a,\\ b)[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{(x,y)\\in\\mathbb{R}^{2},|,(x-a)^{2}+(y-b)^{2}<\\delta^{2}}[\/latex]<\/p>\n<\/section>\n<figure id=\"attachment_970\" aria-describedby=\"caption-attachment-970\" style=\"width: 423px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-970\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28223402\/4-2-1.jpeg\" alt=\"On the xy plane, the point (2, 1) is shown, which is the center of a circle of radius \u03b4.\" width=\"423\" height=\"197\" \/><figcaption id=\"caption-attachment-970\" class=\"wp-caption-text\">Figure 1. A [latex]\\delta[\/latex] disk centered around the point [latex](2,1)[\/latex].<\/figcaption><\/figure>\n<p class=\"whitespace-normal break-words\">The idea of a [latex]\\delta[\/latex] disk appears in the definition of the limit of a function of two variables. If [latex]\\delta[\/latex] is small, then all the points [latex](x,\\ y)[\/latex] in the [latex]\\delta[\/latex] disk are close to [latex](a,\\ b)[\/latex]. This is completely analogous to [latex]x[\/latex] being close to [latex]a[\/latex] in the definition of a limit of a function of one variable. In one dimension, we express this restriction as<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]a-\\delta<x<a+\\delta[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">In more than one dimension, we use a [latex]\\delta[\/latex] disk.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>limit of a function of two variables<\/h3>\n<p class=\"whitespace-normal break-words\">Let [latex]f[\/latex] be a function of two variables. The <strong>limit<\/strong> of [latex]f(x,y)[\/latex] as [latex](x,\\ y)[\/latex] approaches [latex](a,\\ b)[\/latex] is [latex]L[\/latex], written:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{(x,y)\\to(a,b)}{\\text{lim}}f(x,y)=L[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">if for each [latex]\\epsilon>0[\/latex] there exists a [latex]\\delta>0[\/latex] such that for all points [latex](x,\\ y)[\/latex] in a [latex]\\delta[\/latex] disk around [latex](a,\\ b)[\/latex], except possibly for [latex](a,\\ b)[\/latex] itself, we have:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]|f(x,y)-L|<\\epsilon[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">In symbols:<\/p>\n<p>&nbsp;<\/p>\n<p class=\"whitespace-normal break-words\">For any [latex]\\varepsilon>0[\/latex], there exists [latex]\\delta>0[\/latex] such that [latex]|f(x,y)-L|<\\varepsilon[\/latex] whenever [latex]0<\\sqrt{(x-a)^{2}+(y-b)^{2}}<\\delta[\/latex]<\/p>\n<div><\/div>\n<\/section>\n<figure id=\"attachment_971\" aria-describedby=\"caption-attachment-971\" style=\"width: 646px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-971\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28223617\/4-2-2.jpeg\" alt=\"In xyz space, a function is drawn with point L. This point L is the center of a circle of radius \u0949, with points L \u00b1 \u0949 marked. On the xy plane, there is a point (a, b) drawn with a circle of radius \u03b4 around it. This is denoted the \u03b4-disk. There are dashed lines up from the \u03b4-disk to make a disk on the function, which is called the image of delta disk. Then there are dashed lines from this disk to the circle around the point L, which is called the \u0949-neighborhood of L.\" width=\"646\" height=\"581\" \/><figcaption id=\"caption-attachment-971\" class=\"wp-caption-text\">Figure 2.\u00a0The limit of a function involving two variables requires that\u00a0[latex]f(x,y)[\/latex] be within\u00a0[latex]\\varepsilon[\/latex] of\u00a0[latex]L[\/latex]\u00a0whenever [latex](x,y)[\/latex]\u00a0is within [latex]\\delta[\/latex] of\u00a0[latex](a,b)[\/latex]. The smaller the value of\u00a0[latex]\\varepsilon[\/latex],\u00a0the smaller the value of\u00a0[latex]\\delta[\/latex].<\/figcaption><\/figure>\n<p>Proving that a limit exists using the definition of a limit of a function of two variables can be challenging. Instead, we use the following theorem, which gives us shortcuts to finding limits. The formulas in this theorem are extensions of the formulas in the limit laws theorem for single-variable functions.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>limit laws for functions of two variables theorem<\/h3>\n<p>Let [latex]\\underset{(x,y)\\to(a,b)}{\\text{lim}}f(x,y)=L[\/latex] and [latex]\\underset{(x,y)\\to(a,b)}{\\text{lim}}g(x,y)=M[\/latex], and let [latex]c[\/latex] be a constant. Then:<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Constant\u00a0Law:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}c=c[\/latex]<\/p>\n<p><strong>Identity Law:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}x} & =\\hfill & {a} \\hfill \\\\ \\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}y} & =\\hfill & {b} \\end{array}[\/latex]<\/p>\n<p><strong>Sum Law:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y)+g\\,(x,\\ y))=L+M[\/latex]<\/p>\n<p><strong>Difference Law:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y)-(g\\,(x,\\ y))=L-M[\/latex]<\/p>\n<p><strong>Constant Multiple<\/strong> <strong>Law:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(cf\\,(x,\\ y))=cL[\/latex]<\/p>\n<p><strong>Product Law:\u00a0<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y)\\,g\\,(x,\\ y))=LM[\/latex]<\/p>\n<p><strong>Quotient Law:\u00a0<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}\\frac{f\\,(x,\\ y)}{g\\,(x,\\ y)}=\\frac{L}{M}[\/latex]<\/p>\n<p style=\"text-align: left;\">for [latex]M\\neq{0}[\/latex]<\/p>\n<p><strong>Power Law:\u00a0<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y))^{n}=L^{n}[\/latex]<\/p>\n<p>for any positive integer [latex]n[\/latex].<\/p>\n<p><strong>Root Law:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}\\sqrt[n]{f\\,(x,\\ y)}=\\sqrt[n]{L}[\/latex]<\/p>\n<p>for all [latex]L[\/latex] if [latex]n[\/latex] is odd and positive, and for [latex]L\\,\\geq\\,0[\/latex] if [latex]n[\/latex] is even and positive.<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">The proofs of these properties are similar to those for the limits of functions of one variable. We can apply these laws to finding limits of various functions.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Find each of the following limits:<\/p>\n<ol>\n<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)[\/latex]<\/li>\n<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}\\frac{2x+3y}{4x-3y}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793933114\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793933114\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>First use the sum and difference laws to separate the terms:<\/li>\n<\/ol>\n<p>[latex]\\hspace{5cm}\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\hspace{5cm}=\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x^{2}\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}2xy\\bigg) +\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y^{2}\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}4x\\bigg) \\\\ \\hspace{5cm}+\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}6\\bigg).[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Next,\u00a0use the constant multiple law on the second, third, fourth, and fifth limits:<\/p>\n<div style=\"text-align: center;\">[latex]\\hspace{5cm}=\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x^{2}\\bigg)-2\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}xy\\bigg)+3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y^{2}\\bigg)-4\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\bigg) \\\\ \\hspace{5cm} +3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}6\\bigg).[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Now, use the power law on the first and third limits, and the product law on the second limit:<\/p>\n<div style=\"text-align: center;\">[latex]\\hspace{5cm} =\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\bigg)^{2}-2\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}xy\\bigg)+3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\bigg)^{2} \\\\ \\hspace{5cm} -4\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\bigg)+3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}6\\bigg).[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Last, use the identity laws on the first six limits and the constant law on the last limit:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill{\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)}& =\\hfill & {(2)^{2}-2(2)(-1)+3(-1)^{2}-4(2)+3(-1)-6} \\hfill \\\\ \\hfill & =\\hfill & {-6} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<ol>\n<li>Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, constant multiple law, and identity law,\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(4x-3y)} & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}4x-\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y}\\hfill \\\\ \\hfill & =\\hfill & {4\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\big)-3\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\big)} \\hfill \\\\ \\hfill & =\\hfill & {4(2)-3(-1)=11.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Since the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of the numerator using the difference law, constant multiple law, and identity law:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(2x+3y)} & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}2x+\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y}\\hfill \\\\ \\hfill & =\\hfill & {2\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\big)+3\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\big)} \\hfill \\\\ \\hfill & =\\hfill & {2(2)+3(-1)}\\hfill \\\\ \\hfill & =\\hfill & {1.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Therefore, according to the quotient law we have<\/p>\n<div style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}\\frac{2x+3y}{4x-3y}=\\frac{\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(2x+3y)}{\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(4x-3y)}=\\frac{1}{11}.[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Since we are taking the limit of a function of two variables, the point [latex](a,\\ b)[\/latex] is in [latex]\\mathbb{R}^{2}[\/latex], and it is possible to approach this point from an infinite number of directions. Sometimes when calculating a limit, the answer varies depending on the path taken toward [latex](a,\\ b)[\/latex]. If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of path taken<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Show that neither of the following limits exist:<\/p>\n<ol>\n<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{2xy}{3x^{2}+y^{2}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1197793933114\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1197793933114\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The domain of the function [latex]f\\,(x,\\ y)=\\frac{2xy}{3x^{2}+y^{2}}[\/latex]\u00a0consists of all points in the [latex]xy[\/latex]-plane\u00a0except for the point [latex](0,\\ 0)[\/latex]\u00a0(Figure 3). To show that the limit does not exist as [latex](x,\\ y)[\/latex] approaches [latex](0,\\ 0)[\/latex],\u00a0we note that it is impossible to satisfy the definition of a limit of a function of two variables because of the fact that the function takes different values along different lines passing through point [latex](0,\\ 0)[\/latex].\u00a0First, consider the line [latex]y=0[\/latex] in the [latex]xy[\/latex]-plane. Substituting\u00a0[latex]y=0[\/latex] into [latex]f\\,(x,\\ y)[\/latex] gives\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ 0)=\\frac{2x(0)}{3x^{2}+0^{2}}=0}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>For any value of [latex]x[\/latex].\u00a0Therefore the value of [latex]f[\/latex]\u00a0remains constant for any point on the [latex]x[\/latex]-axis, and as [latex]y[\/latex] approaches zero, the function remains fixed at zero.\u00a0Next, consider the line [latex]y=x[\/latex]. Substituting [latex]y=x[\/latex] into [latex]f\\,(x,\\ y)[\/latex] gives<\/p>\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ x)=\\frac{2x(x)}{3x^{2}+x^{2}}=\\frac{2x^{2}}{4x^{2}}=\\frac{1}{2}}.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>This is true for any point on the line [latex]y=x[\/latex]. If we let [latex]x[\/latex] approach zero\u00a0while staying on this line, the value of the function remains fixed at [latex]\\frac{1}{2}[\/latex],\u00a0regardless of how small [latex]x[\/latex] is.\u00a0Choose a value for [latex]\\epsilon[\/latex] that is less than [latex]1\/2-[\/latex]say, [latex]1\/4[\/latex]. Then, no matter how small a [latex]\\delta[\/latex] disk\u00a0we draw around [latex](0,\\ 0)[\/latex], the values of [latex]f\\,(x,\\ y)[\/latex] for points inside that [latex]\\delta[\/latex] disk will include both [latex]0[\/latex] and [latex]\\frac{1}{2}.[\/latex] Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.<\/p>\n<figure id=\"attachment_973\" aria-describedby=\"caption-attachment-973\" style=\"width: 451px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-973\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28224845\/4-2-3.jpeg\" alt=\"In xyz space, the function f(x, y) = 2xy\/(3x2 + y2) is shown, which is a slightly twisted plane, with values of 0 along the line y = 0 and values of \u00bd along the line y = x.\" width=\"451\" height=\"288\" \/><figcaption id=\"caption-attachment-973\" class=\"wp-caption-text\">Figure 3. Graph of the function [latex]\\small{f(x,y)=(2xy)\/(3x^{2}+y^{2})}[\/latex]. Along the line [latex]\\small{y=0}[\/latex], the function is equal to zero; along the line [latex]\\small{y=x}[\/latex], the function is equal to [latex]\\small{\\frac{1}{2}}[\/latex].<\/figcaption><\/figure>\n<\/li>\n<li>In a similar fashion to 1., we can approach the origin along any straight line passing through the origin. If we try the [latex]x[\/latex]-axis (i.e., [latex]y=0[\/latex]),\u00a0then the function remains fixed at zero. The same is true for the [latex]y[\/latex]-axis.\u00a0Suppose we approach the origin along a straight line of slope [latex]k[\/latex]\u00a0The equation of this line is [latex]y=kx[\/latex]. Then the limit becomes\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}} & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4x(kx)^{2}}{x^{2}+3(kx)^{4}}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4k^{2}x^{3}}{x^{2}+3k^{4}x^{4}}} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4k^{2}x}{1+3k^{4}x^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}4k^{2}x}{\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}(1+3k^{4}x^{2})}}\\hfill \\\\ \\hfill & =\\hfill & {0}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Regardless of the value of [latex]k[\/latex].\u00a0It would seem that the limit is equal to zero. What if we chose a curve passing through the origin instead? For example, we can consider the parabola given by the equation [latex]x=y^{2}[\/latex]. Substituting [latex]y^{2}[\/latex] in place of [latex]x[\/latex] in [latex]f\\,(x,\\ y)[\/latex] gives<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}} & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4(y^{2})y^{2}}{(y^{2})^{2}+3y^{4}}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4y^{4}}{y^{4}+3y^{4}}} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}1}\\hfill \\\\ \\hfill & =\\hfill & {1.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>By the same logic\u00a0in 1., it is impossible to find a [latex]\\delta[\/latex] disk\u00a0around the origin that satisfies the definition of the limit for any value of [latex]\\epsilon\\,<\\,1[\/latex]. Therefore, [latex]\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}[\/latex] does not exist.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":11,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-596","chapter","type-chapter","status-publish","hentry"],"part":426,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/596","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/596\/revisions"}],"predecessor-version":[{"id":1737,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/596\/revisions\/1737"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/426"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/596\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=596"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=596"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=596"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=596"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}