Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

1.4E: Exercises

  • Page ID
    10660
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    Exercise \(\PageIndex{1}\): Limit at infinity

    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:

    \( \displaystyle = 3(0)/ 2\)

    \( \displaystyle = 0\)

    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:

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

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

    \( \displaystyle = 1\)

    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:

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

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

    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

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

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

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

    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:

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

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

    \( \displaystyle 2\)

    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:

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

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

    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:

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

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

    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:

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

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

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

    Answer

    \( \displaystyle = 0\)

    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:

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

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

    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 \(\PageIndex{2}\): 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 and Attributions

    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)