4.6E: Exercises
- Page ID
- 10660
This page is a draft and is under active development.
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercise \(\PageIndex{1}\): Limit at infinity
1. Find \( \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x\)
- Answer
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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
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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
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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
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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
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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
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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
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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
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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
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\( \displaystyle = 0\)
10.Find \( \displaystyle \lim_{x \to \infty} \frac{3x^2-2x}{5x^2+1}\)
- Answer
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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
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\(\displaystyle\lim_{y \to -\infty} \frac{\sqrt{7y^2+6y-8}}{5-y}= \displaystyle\lim_{y \to -\infty} \frac{\sqrt{7y^2}}{-y}\)
\(\displaystyle\lim_{y \to -\infty} \frac{\frac{\sqrt{7y^2}}{|y|}}{\frac{-y}{|y|}}= \displaystyle\lim_{y \to -\infty} \frac{\frac{\sqrt{7y^2}}{\sqrt{y^2}}}{\frac{-y}{-y}} = \sqrt{7} \)
Exercise \(\PageIndex{2}\): Horizontal Asymptotes
Find horizontal asymptote(s) of the function \[ y= \displaystyle \frac{5+3x}{ \sqrt{x^2+x-4}}.\]
- Answer
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\(y=3\) 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)