{"id":3083,"date":"2024-06-12T16:46:06","date_gmt":"2024-06-12T16:46:06","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&p=3083"},"modified":"2024-08-05T13:17:04","modified_gmt":"2024-08-05T13:17:04","slug":"limits-at-infinity-and-asymptotes-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/limits-at-infinity-and-asymptotes-learn-it-2\/","title":{"raw":"Limits at Infinity and Asymptotes: Learn It 2","rendered":"Limits at Infinity and Asymptotes: Learn It 2"},"content":{"raw":"

Limits at Infinity Cont.<\/h2>\r\n

Formal Definitions<\/h3>\r\n

Earlier, we used the terms arbitrarily close<\/em>, arbitrarily large<\/em>, and sufficiently large<\/em> to define limits at infinity informally. Although these terms provide accurate descriptions of limits at infinity, they are not precise mathematically.<\/p>\r\n

Here are more formal definitions of limits at infinity.\u00a0<\/p>\r\n

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limits at infinity (formal)<\/h3>\r\n<\/div>\r\n

We say a function [latex]f[\/latex] has a limit at infinity, if there exists a real number [latex]L[\/latex] such that for all [latex]\\varepsilon >0[\/latex], there exists [latex]N>0[\/latex] such that<\/p>\r\n

[latex]|f(x)-L|<\\varepsilon [\/latex]<\/div>\r\n

for all [latex]x>N[\/latex]. In that case, we write<\/p>\r\n

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

We say a function [latex]f[\/latex] has a limit at negative infinity if there exists a real number [latex]L[\/latex] such that for all [latex]\\varepsilon >0[\/latex], there exists [latex]N<0[\/latex] such that<\/p>\r\n

[latex]|f(x)-L|<\\varepsilon [\/latex]<\/div>\r\n

for all [latex]x<N[\/latex]. In that case, we write<\/p>\r\n

[latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)=L[\/latex]<\/div>\r\n<\/section>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"369\"]\"The Figure 9. For a function with a limit at infinity, for all [latex]x>N[\/latex], [latex]|f(x)-L|<\\varepsilon [\/latex].[\/caption]\r\n\r\n

Earlier in this section, we used graphical evidence and numerical evidence to conclude that [latex]\\underset{x\\to \\infty }{\\lim}\\left(2+\\frac{1}{x}\\right)=2[\/latex]. Here we use the formal definition of limit at infinity to prove this result.<\/p>\r\n

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Use the formal definition of limit at infinity to prove that [latex]\\underset{x\\to \\infty }{\\lim}\\left(2+\\frac{1}{x}\\right)=2[\/latex].<\/p>\r\n

[reveal-answer q=\"fs-id1165042369578\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1165042369578\"]<\/p>\r\n

Let [latex]\\varepsilon >0[\/latex]. Let [latex]N=\\frac{1}{\\varepsilon }[\/latex].<\/p>\r\n

Therefore, for all [latex]x>N[\/latex], we have:<\/p>\r\n

[latex]\\left| 2+\\dfrac{1}{x}-2 \\right| =\\left| \\dfrac{1}{x} \\right|=\\dfrac{1}{x}<\\dfrac{1}{N}=\\varepsilon[\/latex].<\/div>\r\n

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