{"id":2472,"date":"2024-05-22T18:41:39","date_gmt":"2024-05-22T18:41:39","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&p=2472"},"modified":"2024-08-05T01:39:22","modified_gmt":"2024-08-05T01:39:22","slug":"the-precise-definition-of-a-limit-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-precise-definition-of-a-limit-learn-it-3\/","title":{"raw":"The Precise Definition of a Limit: Learn It 3","rendered":"The Precise Definition of a Limit: Learn It 3"},"content":{"raw":"

One-Sided and Infinite Limits<\/h2>\r\n

One-Sided Limits<\/h3>\r\n

After gaining an intuitive understanding of limits and moving to a more rigorous definition, we now revisit one-sided limits. We modify the epsilon-delta definition of a limit to give formal definitions for limits from the right and left at a point. These definitions only require slight modifications from the standard limit definition. In the definition of the limit from the right, the inequality [latex]0 < x-a < \\delta[\/latex] replaces [latex]0 < |x-a| < \\delta[\/latex], which ensures that we only consider values of [latex]x[\/latex] that are greater than (to the right of) [latex]a[\/latex]. Similarly, in the definition of the limit from the left, the inequality [latex]-\\delta < x-a < 0[\/latex] replaces [latex]0 < |x-a| < \\delta[\/latex], which ensures that we only consider values of [latex]x[\/latex] that are less than (to the left of) [latex]a[\/latex].<\/p>\r\n

\r\n

one-sided limits definitions<\/h3>\r\n

Limit from the Right:<\/strong> Let [latex]f(x)[\/latex] be defined over an open interval of the form [latex](a,b)[\/latex] where [latex]a < b[\/latex]. Then,<\/p>\r\n

[latex]\\underset{x\\to a^+}{\\lim}f(x)=L[\/latex]<\/div>\r\n
\u00a0<\/div>\r\n

if for every [latex]\\varepsilon >0[\/latex], there exists a [latex]\\delta >0[\/latex] such that if [latex]0 < x-a < \\delta[\/latex], then [latex]|f(x)-L|<\\varepsilon[\/latex].<\/p>\r\n

 <\/p>\r\n

Limit from the Left:<\/strong> Let [latex]f(x)[\/latex] be defined over an open interval of the form [latex](a,b)[\/latex] where [latex]a < b[\/latex]. Then,<\/p>\r\n

[latex]\\underset{x\\to b^-}{\\lim}f(x)=L[\/latex]<\/div>\r\n
\u00a0<\/div>\r\n

if for every [latex]\\varepsilon >0[\/latex], there exists a [latex]\\delta >0[\/latex] such that if [latex]0 < b-x < \\delta[\/latex], then [latex]|f(x)-L|<\\varepsilon[\/latex].<\/p>\r\n<\/section>\r\n

\r\n

Prove that [latex]\\underset{x\\to 4^+}{\\lim}\\sqrt{x-4}=0[\/latex].<\/p>\r\n

[reveal-answer q=\"fs-id1170572601349\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1170572601349\"]\r\n\r\n

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

Choose [latex]\\delta =\\varepsilon^2[\/latex]. Since we ultimately want [latex]|\\sqrt{x-4}-0|<\\varepsilon[\/latex], we manipulate this inequality to get [latex]\\sqrt{x-4}<\\varepsilon[\/latex] or, equivalently, [latex]0 < x-4 < \\varepsilon^2[\/latex], making [latex]\\delta =\\varepsilon^2[\/latex] a clear choice. \""A\"<\/p>\r\n

Assume [latex]0 < x-4 < \\delta[\/latex]. Thus, [latex]0 < x-4 < \\varepsilon^2[\/latex]. Hence, [latex]0<\\sqrt{x-4}<\\varepsilon[\/latex]. Finally, [latex]|\\sqrt{x-4}-0|<\\varepsilon[\/latex].<\/p>\r\n

Therefore, [latex]\\underset{x\\to 4^+}{\\lim}\\sqrt{x-4}=0[\/latex].<\/p>\r\n

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

https:\/\/www.youtube.com\/watch?v=5_q7_Zx26RY<\/p>\r\n

Closed Captioning and Transcript Information for Video<\/p>\r\n

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n

You can view the transcript for this segmented clip of \"2.5 Precise Definition of a Limit\" here (opens in new window)<\/a>.<\/p>\r\n

[\/hidden-answer]<\/p>\r\n<\/div>\r\n<\/section>\r\n

Infinite Limits<\/h3>\r\n

To understand infinite limits, we look at how functions behave as [latex]x[\/latex] approaches a certain value [latex]a[\/latex]. For [latex]\\underset{x\\to a}{\\lim}f(x)=+\\infty[\/latex], we want [latex]f(x)[\/latex] to get arbitrarily large as [latex]x[\/latex] approaches [latex]a[\/latex]. Instead of the requirement that [latex]|f(x)-L|<\\varepsilon[\/latex] for arbitrarily small [latex]\\varepsilon[\/latex] when [latex]0<|x-a|<\\delta[\/latex] for small enough [latex]\\delta[\/latex], we want [latex]f(x)>M[\/latex] for arbitrarily large positive [latex]M[\/latex] when [latex]0<|x-a|<\\delta[\/latex] for small enough [latex]\\delta[\/latex]. The figure below illustrates this idea by showing the value of [latex]\\delta[\/latex] for successively larger values of [latex]M[\/latex].<\/p>\r\n

\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"Two Figure 6. These graphs plot values of [latex]\\delta[\/latex] for [latex]M[\/latex] to show that [latex]\\underset{x\\to a}{\\lim}f(x)=+\\infty[\/latex].[\/caption]\r\n<\/div>\r\n
\r\n

infinite limit definition<\/h3>\r\n

Let [latex]f(x)[\/latex] be defined for all [latex]x\\ne a[\/latex] in an open interval containing [latex]a[\/latex]. Then, we have an infinite limit<\/p>\r\n

[latex]\\underset{x\\to a}{\\lim}f(x)=+\\infty[\/latex]<\/div>\r\n
\u00a0<\/div>\r\n

if for every [latex]M>0[\/latex], there exists [latex]\\delta >0[\/latex] such that if [latex]0<|x-a|<\\delta[\/latex], then [latex]f(x)>M[\/latex].<\/p>\r\n

 <\/p>\r\n

Let [latex]f(x)[\/latex] be defined for all [latex]x\\ne a[\/latex] in an open interval containing [latex]a[\/latex]. Then, we have a negative infinite limit<\/p>\r\n

[latex]\\underset{x\\to a}{\\lim}f(x)=\u2212\\infty[\/latex]<\/div>\r\n
\u00a0<\/div>\r\n

if for every [latex]M>0[\/latex], there exists [latex]\\delta >0[\/latex] such that if [latex]0<|x-a|<\\delta[\/latex], then [latex]f(x)<\u2212M[\/latex].<\/p>\r\n<\/section>","rendered":"

One-Sided and Infinite Limits<\/h2>\n

One-Sided Limits<\/h3>\n

After gaining an intuitive understanding of limits and moving to a more rigorous definition, we now revisit one-sided limits. We modify the epsilon-delta definition of a limit to give formal definitions for limits from the right and left at a point. These definitions only require slight modifications from the standard limit definition. In the definition of the limit from the right, the inequality [latex]0 < x-a < \\delta[\/latex] replaces [latex]0 < |x-a| < \\delta[\/latex], which ensures that we only consider values of [latex]x[\/latex] that are greater than (to the right of) [latex]a[\/latex]. Similarly, in the definition of the limit from the left, the inequality [latex]-\\delta < x-a < 0[\/latex] replaces [latex]0 < |x-a| < \\delta[\/latex], which ensures that we only consider values of [latex]x[\/latex] that are less than (to the left of) [latex]a[\/latex].<\/p>\n

\n

one-sided limits definitions<\/h3>\n

Limit from the Right:<\/strong> Let [latex]f(x)[\/latex] be defined over an open interval of the form [latex](a,b)[\/latex] where [latex]a < b[\/latex]. Then,<\/p>\n

[latex]\\underset{x\\to a^+}{\\lim}f(x)=L[\/latex]<\/div>\n
\u00a0<\/div>\n

if for every [latex]\\varepsilon >0[\/latex], there exists a [latex]\\delta >0[\/latex] such that if [latex]0 < x-a < \\delta[\/latex], then [latex]|f(x)-L|<\\varepsilon[\/latex].<\/p>\n

 <\/p>\n

Limit from the Left:<\/strong> Let [latex]f(x)[\/latex] be defined over an open interval of the form [latex](a,b)[\/latex] where [latex]a < b[\/latex]. Then,<\/p>\n

[latex]\\underset{x\\to b^-}{\\lim}f(x)=L[\/latex]<\/div>\n
\u00a0<\/div>\n

if for every [latex]\\varepsilon >0[\/latex], there exists a [latex]\\delta >0[\/latex] such that if [latex]0 < b-x < \\delta[\/latex], then [latex]|f(x)-L|<\\varepsilon[\/latex].<\/p>\n<\/section>\n

\n

Prove that [latex]\\underset{x\\to 4^+}{\\lim}\\sqrt{x-4}=0[\/latex].<\/p>\n

\n