Skip to main content
Mathematics LibreTexts

2.5E: Exercises for Section 2.5

  • Page ID
    50132
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    For exercises 1 - 5, examine the graphs. Identify where the vertical asymptotes are located.

    1)

    The function graphed decreases very rapidly as it approaches x = 1 from the left, and on the other side of x = 1, it seems to start near infinity and then decrease rapidly.

    Answer
    \(x=1\)

    2)

    The function graphed increases very rapidly as it approaches x = −3 from the left, and on the other side of x = −3, it seems to start near negative infinity and then increase rapidly to form a sort of U shape that is pointing down, with the other side of the U being at x = 2. On the other side of x = 2, the graph seems to start near infinity and then decrease rapidly.

    3)

    The function graphed decreases very rapidly as it approaches x = −1 from the left, and on the other side of x = −1, it seems to start near negative infinity and then increase rapidly to form a sort of U shape that is pointing down, with the other side of the U being at x = 2. On the other side of x = 2, the graph seems to start near infinity and then decrease rapidly.

    Answer
    \(x=−1,\;x=2\)

    4)

    The function graphed decreases very rapidly as it approaches x = 0 from the left, and on the other side of x = 0, it seems to start near infinity and then decrease rapidly to form a sort of U shape that is pointing up, with the other side of the U being at x = 1. On the other side of x = 1, there is another U shape pointing down, with its other side being at x = 2. On the other side of x = 2, the graph seems to start near negative infinity and then increase rapidly.

    5)

    The function graphed decreases very rapidly as it approaches x = 0 from the left, and on the other side of x = 0, it seems to start near infinity and then decrease rapidly to form a sort of U shape that is pointing up, with the other side being a normal function that appears as if it will take the entirety of the values of the x-axis.

    Answer
    \(x=0\)

     

    For the functions \(f(x)\) in exercises 6 - 10, determine whether there is an asymptote at \(x=a\). Justify your answer without graphing on a calculator.

    6) \(f(x)=\dfrac{x+1}{x^2+5x+4},\quad a=−1\)

    7) \(f(x)=\dfrac{x}{x−2},\quad a=2\)

    Answer
    Yes, there is a vertical asymptote at \(x = 2\).

    8) \(f(x)=(x+2)^{3/2},\quad a=−2\)

    9) \(f(x)=(x−1)^{−1/3},\quad a=1\)

    Answer
    Yes, there is vertical asymptote at \(x = 1\).

    10) \(f(x)=1+x^{−2/5},\quad a=1\)

     

    In exercises 11 - 20, evaluate the limit.

    11) \(\displaystyle \lim_{x→∞}\frac{1}{3x+6}\)

    Answer
    \(\displaystyle \lim_{x→∞}\frac{1}{3x+6} = 0\)

    12) \(\displaystyle \lim_{x→∞}\frac{2x−5}{4x}\)

    13) \(\displaystyle \lim_{x→∞}\frac{x^2−2x+5}{x+2}\)

    Answer
    \(\displaystyle \lim_{x→∞}\frac{x^2−2x+5}{x+2} = ∞\)

    14) \(\displaystyle \lim_{x→−∞}\frac{3x^3−2x}{x^2+2x+8}\)

    15) \(\displaystyle \lim_{x→−∞}\frac{x^4−4x^3+1}{2−2x^2−7x^4}\)

    Answer
    \(\displaystyle \lim_{x→−∞}\frac{x^4−4x^3+1}{2−2x^2−7x^4} = −\frac{1}{7}\)

    16) \(\displaystyle \lim_{x→∞}\frac{3x}{\sqrt{x^2+1}}\)

    17) \(\displaystyle \lim_{x→−∞}\frac{\sqrt{4x^2−1}}{x+2}\)

    Answer
    \(\displaystyle \lim_{x→−∞}\frac{\sqrt{4x^2−1}}{x+2} = -2\)

    18) \(\displaystyle \lim_{x→∞}\frac{4x}{\sqrt{x^2−1}}\)

    19) \(\displaystyle \lim_{x→−∞}\frac{4x}{\sqrt{x^2−1}}\)

    Answer
    \(\displaystyle \lim_{x→−∞}\frac{4x}{\sqrt{x^2−1}} = -4\)

    20) \(\displaystyle \lim_{x→∞}\frac{2\sqrt{x}}{x−\sqrt{x}+1}\)

     

    For exercises 21 - 25, find the horizontal and vertical asymptotes.

    21) \(f(x)=x−\dfrac{9}{x}\)

    Answer
    Horizontal: none,
    Vertical: \(x=0\)

    22) \(f(x)=\dfrac{1}{1−x^2}\)

    23) \(f(x)=\dfrac{x^3}{4−x^2}\)

    Answer
    Horizontal: none,
    Vertical: \(x=±2\)

    24) \(f(x)=\dfrac{x^2+3}{x^2+1}\)

    25) \(f(x)=\sin(x)\sin(2x)\)

    Answer
    Horizontal: none,
    Vertical: none

    26) \(f(x)=\cos x+\cos(3x)+\cos(5x)\)

    27) \(f(x)=\dfrac{x\sin(x)}{x^2−1}\)

    Answer
    Horizontal: \(y=0,\)
    Vertical: \(x=±1\)

    28) \(f(x)=\dfrac{x}{\sin(x)}\)

    29) \(f(x)=\dfrac{1}{x^3+x^2}\)

    Answer
    Horizontal: \(y=0,\)
    Vertical: \(x=0\) and \(x=−1\)

    30) \(f(x)=\dfrac{1}{x−1}−2x\)

    31) \(f(x)=\dfrac{x^3+1}{x^3−1}\)

    Answer
    Horizontal: \(y=1,\)
    Vertical: \(x=1\)

    32) \(f(x)=\dfrac{\sin x+\cos x}{\sin x−\cos x}\)

    33) \(f(x)=x−\sin x\)

    Answer
    Horizontal: none,
    Vertical: none

    34) \(f(x)=\dfrac{1}{x}−\sqrt{x}\)

     

    For exercises 35 - 38, construct a function \(f(x)\) that has the given asymptotes.

    35) \(x=1\) and \(y=2\)

    Answer
    Answers will vary, for example: \(y=\dfrac{2x}{x−1}\)

    36) \(x=1\) and \(y=0\)

    37) \(y=4, \;x=−1\)

    Answer
    Answers will vary, for example: \(y=\dfrac{4x}{x+1}\)

    38) \(x=0\)

     

    In exercises 39 - 43, graph the function on a graphing calculator on the window \(x=[−5,5]\) and estimate the horizontal asymptote or limit. Then, calculate the actual horizontal asymptote or limit.

    39) [T] \(f(x)=\dfrac{1}{x+10}\)

    Answer
    \(\displaystyle \lim_{x→∞}\frac{1}{x+10}=0\)  so \(f\) has a horizontal asymptote of \(y=0\).

    40) [T] \(f(x)=\dfrac{x+1}{x^2+7x+6}\)

    41) [T] \(\displaystyle \lim_{x→−∞}x^2+10x+25\)

    Answer
    \(\displaystyle \lim_{x→−∞}x^2+10x+25 = ∞\)

    42) [T] \(\displaystyle \lim_{x→−∞}\frac{x+2}{x^2+7x+6}\)

    43) [T] \(\displaystyle \lim_{x→∞}\frac{3x+2}{x+5}\)

    Answer
    \(\displaystyle \lim_{x→∞}\frac{3x+2}{x+5}=3\)  so this function has a horizontal asymptote of \(y=3\).

     

    Contributors and Attributions

    • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


    This page titled 2.5E: Exercises for Section 2.5 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.

    • Was this article helpful?