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Mathematics LibreTexts

1.4E Exercises

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    10660
  • [ "stage:draft", "article:topic" ]

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    Limit at infinity

    Exercise \(\PageIndex{1}\)

    Find \( \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x\)

    Answer

    To find  \( \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x\), First multiply by the conjugate

     \( \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x =  \lim_{x \to \infty} \frac{(\sqrt{x^2+3}-x)(\sqrt{x^2+3}+x)}{\sqrt{x^2+3}+x}\)

    Then reduce the numerator

     \( \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x = \lim_{x \to \infty} \frac{3}{\sqrt{x^2+3}+x}\)

    Now divide by \( \sqrt{x^2}=|x|\), 

     \( \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x = \lim_{x \to \infty} \frac{3/|x|}{(\sqrt{x^2+3}+x)/ \sqrt{x^2}}\)

    Since  \(x \to \infty\),  replace \(|x|\) by \(x\).

    Thus,  \( \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x = \lim_{x \to \infty} \frac{3/x}{(\sqrt{1+3/\sqrt{x^2}}+x/x)

    This result will be:

    \( \displaystyle = 3(0)/ 2\)

    \( \displaystyle = 0\)

    Exercise \(\PageIndex{2}\)

    Find \( \displaystyle \lim_{x \to - \infty} \frac{\sqrt{x^4+6x}}{x^2-6}\)

    Answer

    Factor out the highest degree variable in the numerator and the denominator, \( \displaystyle \sqrt{x^4}=|x^2|=x^2\).

    \( \displaystyle \lim_{x \to - \infty} \frac{\sqrt{x^4+6x}}{x^2-6} = \lim_\limits{x \to - \infty} \frac{x^2\sqrt{1 + \frac{6x}{x^3}}}{(x^2)(1- \frac{6}{x^2})}\)

     

    Evaluate:

    \( \displaystyle = \frac{\sqrt{1+0}}{1-0}\)

    \( \displaystyle = \frac{1}{1}\)

    \( \displaystyle = 1\) 

    Exercise \(\PageIndex{3}\)

    Find \( \displaystyle \lim\limits_{t \to \infty} \frac{7-t}{\sqrt{2+2t^2}}\)

    Answer

    Factor out the highest degree variable in the numerator and denominator

    = \( \displaystyle \lim\limits_{t \to\infty} \frac{t (\frac{7}{t} -1)}{|t|\sqrt{\frac{2}{t^2}+2}}\)

    Evaluate:

    \( \displaystyle = \frac{0-1}{\sqrt{0+2}}\)

    \( \displaystyle = \frac{-1}{\sqrt{2}}\)

    Exercise \(\PageIndex{4}\)

    Find \( \displaystyle \lim_{t \to - \infty} \frac{7-t}{\sqrt{2+2t^2}}\)

    Answer

    Factor out the highest degree variable in the numerator and denominator

    \( \displaystyle =\lim_{t \to \infty} \frac{(t)(\frac{7}{t}-1)}{|t|\sqrt{\frac{2}{t^2}+2}}\)

    \( \displaystyle = \frac{0-1}{(-1)\sqrt{0+2}}\)

    \( \displaystyle = \frac{1}{\sqrt{2}}\)

    Exercise \(\PageIndex{5}\)

    Find \( \displaystyle \lim_{y \to - \infty} \frac{\sqrt{4y^2-2}}{2-y}\)

    Answer

    Factor out the highest degree variable in the numerator and denominator

    \( \displaystyle = \lim_{y \to -\infty} \frac{y\sqrt{4-\frac{2}{y^2}}}{|y|(\frac{2}{y}-1)}\)

    When \(y \to -\infty \)  \(   |y | \)approaches to \(-\infty\).

    Evaluate:

    \( \displaystyle \frac{\sqrt{4-0}}{-(0-1)}\)

    \( \displaystyle \frac{2}{1}\)

    \( \displaystyle 2\)

    Exercise \(\PageIndex{6}\)

    Find \( \displaystyle \lim_{s \to \infty} \sqrt[5]{ \frac{5s^8 - 4s^3}{2s^8-1}}\)

    Answer

    Because the limit is approaching infinity, only the highest degree polynomials need to be considered. This then simplifies the limit to:

    = \( \displaystyle \lim_{s \to \infty} \sqrt[5]{\frac{5s^8}{2s^8}}\)

    \(= \displaystyle \sqrt[5]{\frac{5}{2}}\)

    Exercise \(\PageIndex{7}\)

    Find \( \displaystyle \lim_{x \to \infty} \sqrt[3]{ \frac{3+2x+5x^2}{1+4x^2}}\)

    Answer

    Because the limit is approaching infinity, only the highest degree polynomials need to be considered. This then simplifies the limit to:

    \( \displaystyle = \lim_{x \to \infty} \sqrt[3]{\frac{5x^2}{4x^2}}\)

    \( \displaystyle = \sqrt[3]{\frac{5}{4}}\).

    Exercise \(\PageIndex{8}\)

    Find \( \displaystyle \lim_{t \to \infty} \frac {11-t^5}{11t^5 +1}\)

    Answer

    Because the limit is approaching infinity, only the highest degree polynomials need to be considered. This then simplifies the limit to:

    \( \displaystyle = \lim_{t \to \infty} \frac {-t^5}{11t^5}\)

    \( \displaystyle = \frac {-1}{11}\)  

    Exercise \(\PageIndex{9}\)

    Find \( \displaystyle  \lim_{x \to -\infty} \frac{3}{x-3}\)

    Answer

    \( \displaystyle = 0\)

    Exercise \(\PageIndex{10}\)

    Find \( \displaystyle \lim_{x \to \infty} \frac{3x^2-2x}{5x^2+1}\)

    Answer

    Because the limit is approaching infinity, only the highest degree polynomials need to be considered. This then simplifies the limit to:

    \( \displaystyle = \lim_{x \to -\infty} \frac{3x^2}{5x^2}\)

    \( \displaystyle = \frac{3}{5}\).

    Exercise \(\PageIndex{11}\)

    Find \(\displaystyle\lim_{y \to -\infty} \frac{\sqrt{7y^2+6y-8}}{5-y}\).

    Answer

    \(\displaystyle\lim_{y \to -\infty} \frac{\sqrt{7y^2+6y-8}}{5-y}= \displaystyle\lim_{y \to -\infty} \frac{\sqrt{7y^2}}{-y}\)

    \(\displaystyle\lim_{y \to -\infty} \frac{\frac{\sqrt{7y^2}}{|y|}}{\frac{-y}{|y|}}= \displaystyle\lim_{y \to -\infty} \frac{\frac{\sqrt{7y^2}}{\sqrt{y^2}}}{\frac{-y}{-y}} = \sqrt{7} \)

    Exercise \(\PageIndex{12}\): Horizontal Asymptotes

    Find horizontal asymptote(s) of the function $$ y= \displaystyle \frac{5+3x}{ \sqrt{x^2+x-4}}.$$

     
    Answer

    \(y=3\) and \(y=-3\).

    Contributors

    Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/

    Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)