{"id":207,"date":"2023-09-20T22:48:16","date_gmt":"2023-09-20T22:48:16","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/proving-limit-laws\/"},"modified":"2025-08-17T22:34:49","modified_gmt":"2025-08-17T22:34:49","slug":"the-precise-definition-of-a-limit-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-precise-definition-of-a-limit-learn-it-2\/","title":{"raw":"The Precise Definition of a Limit: Learn It 2","rendered":"The Precise Definition of a Limit: Learn It 2"},"content":{"raw":"

Advanced Applications of the Epsilon-Delta Definition: Proofs, Non-Existence, and Algebraic Calculations<\/h2>\r\n

Using the Epsilon-Delta Definition of Limits<\/h3>\r\n

We will demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws. The triangle inequality is a key component of this proof, so let's review it first.<\/p>\r\n

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triangle inequality<\/strong><\/h3>\r\n

The triangle inequality<\/strong> states that if [latex]a[\/latex] and [latex]b[\/latex] are any real numbers, then [latex]|a+b|\\le |a|+|b|[\/latex].<\/p>\r\n<\/section>\r\n

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Proof<\/strong><\/p>\r\n

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We prove the following limit law: If [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex] and [latex]\\underset{x\\to a}{\\lim}g(x)=M[\/latex], then [latex]\\underset{x\\to a}{\\lim}(f(x)+g(x))=L+M[\/latex].<\/p>\r\n

Let [latex]\\varepsilon >0[\/latex].<\/p>\r\n

Choose [latex]\\delta_1>0[\/latex] so that if [latex]0<|x-a|<\\delta_1[\/latex], then [latex]|f(x)-L|<\\varepsilon\/2[\/latex].<\/p>\r\n

Choose [latex]\\delta_2>0[\/latex] so that if [latex]0<|x-a|<\\delta_2[\/latex], then [latex]|g(x)-M|<\\varepsilon\/2[\/latex].<\/p>\r\n

Choose [latex]\\delta =\\text{min}\\{\\delta_1,\\delta_2\\}[\/latex].<\/p>\r\n

Assume [latex]0<|x-a|<\\delta[\/latex].<\/p>\r\n

Thus,<\/p>\r\n

[latex]0<|x-a|<\\delta_1[\/latex] and [latex]0<|x-a|<\\delta_2[\/latex]<\/div>\r\n

Hence,<\/p>\r\n

[latex]\\begin{array}{ll} |(f(x)+g(x))-(L+M)| & =|(f(x)-L)+(g(x)-M)| \\\\ & \\le |f(x)-L|+|g(x)-M| \\\\ & <\\dfrac{\\varepsilon}{2}+\\dfrac{\\varepsilon}{2}=\\varepsilon \\end{array}[\/latex]<\/div>\r\n\r\n[latex]\\blacksquare[\/latex]<\/section>\r\n

Exploring the Non-Existence of Limits<\/h3>\r\n

We now explore what it means for a limit not to exist. The limit [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] does not exist if there is no real number [latex]L[\/latex] for which [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex]. For all real numbers [latex]L[\/latex], [latex]\\underset{x\\to a}{\\lim}f(x)\\ne L[\/latex].<\/p>\r\n

To understand what this means, we look at each part of the definition of [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex] together with its opposite.\u00a0<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Translation of the Definition of [latex]\\underset{x\\to a}{\\lim}f(x)=L[\/latex] and its Opposite<\/caption>\r\n
Definition<\/strong><\/th>\r\nOpposite<\/strong><\/th>\r\n<\/tr>\r\n<\/thead>\r\n
1. For every [latex]\\varepsilon >0[\/latex],<\/td>\r\n1. There exists [latex]\\varepsilon >0[\/latex] so that<\/td>\r\n<\/tr>\r\n
2. there exists a [latex]\\delta >0[\/latex] so that<\/td>\r\n2. for every [latex]\\delta >0[\/latex],<\/td>\r\n<\/tr>\r\n
3. if [latex]0<|x-a|<\\delta[\/latex], then [latex]|f(x)-L|<\\varepsilon[\/latex].<\/td>\r\n3. There is an [latex]x[\/latex] satisfying [latex]0<|x-a|<\\delta [\/latex] so that [latex]|f(x)-L|\\ge \\varepsilon[\/latex].<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

Finally, we may state what it means for a limit not to exist. The limit [latex]\\underset{x\\to a}{\\lim}f(x)[\/latex] does not exist if for every real number [latex]L[\/latex], there exists a real number [latex]\\varepsilon >0[\/latex] so that for all [latex]\\delta >0[\/latex], there is an [latex]x[\/latex] satisfying [latex]0<|x-a|<\\delta[\/latex], so that [latex]|f(x)-L|\\ge \\varepsilon[\/latex].<\/p>\r\n

Let\u2019s apply this in the example to show that a limit does not exist.<\/p>\r\n

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Show that [latex]\\underset{x\\to 0}{\\lim}\\dfrac{|x|}{x}[\/latex] does not exist. The graph of [latex]f(x)=\\dfrac{|x|}{x}[\/latex] is shown here:<\/p>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"342\"]\""A\" Graph of f(x) showing discontinuity at x = 0[\/caption]\r\n\r\n

[reveal-answer q=\"fs-id1170572601135\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1170572601135\"]<\/p>\r\n

Suppose that [latex]L[\/latex] is a candidate for a limit. Choose [latex]\\varepsilon =\\frac{1}{2}[\/latex].<\/p>\r\n

Let [latex]\\delta >0[\/latex]. Either [latex]L\\ge 0[\/latex] or [latex]L<0[\/latex]. If [latex]L\\ge 0[\/latex], then let [latex]x=-\\delta\/2[\/latex]. Thus,<\/p>\r\n

[latex]|x-0|=|-\\frac{\\delta}{2}-0|=\\frac{\\delta}{2}<\\delta [\/latex]<\/div>\r\n

and<\/p>\r\n

[latex]|\\frac{|-\\frac{\\delta}{2}|}{-\\frac{\\delta}{2}}-L|=|-1-L|=L+1\\ge 1>\\frac{1}{2}=\\varepsilon[\/latex]<\/div>\r\n

On the other hand, if [latex]L<0[\/latex], then let [latex]x=\\delta\/2[\/latex]. Thus,<\/p>\r\n

[latex]|x-0|=|\\frac{\\delta}{2}-0|=\\frac{\\delta}{2}<\\delta [\/latex]<\/div>\r\n

and<\/p>\r\n

[latex]|\\frac{|\\frac{\\delta}{2}|}{\\frac{\\delta}{2}}-L|=|1-L|=|L|+1\\ge 1>\\frac{1}{2}=\\varepsilon[\/latex]<\/div>\r\n

Thus, for any value of [latex]L[\/latex], [latex]\\underset{x\\to 0}{\\lim}\\frac{|x|}{x}\\ne L[\/latex].<\/p>\r\n

[caption]Watch the following video to see the worked solution to this example. [\/caption]<\/p>\r\n