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Mathematics LibreTexts

1.3E: Exercises

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    Exercise \(\PageIndex{1}\)e:

    Using:

    \[\begin{align}\lim\limits_{x\to9}f(x)=6 \quad \lim\limits_{x\to6}f(x)=9 \end{align}\]
    \[\begin{align} \lim\limits_{x\to9}g(x)=3 \quad \lim\limits_{x\to6}g(x)=3 \end{align}\]

    evaluate the limits given in Exercises 6-13, where possible. If it is not possible to know, state so.

    1. \(\displaystyle \lim\limits_{x\to9}(f(x)+g(x))\)
    2. \(\displaystyle \lim\limits_{x\to9}(3f(x)/g(x))\)
    3. \(\displaystyle \lim\limits_{x\to9} \left ( \frac{f(x)-2g(x)}{g(x)}\right )\)
    4. \(\displaystyle \lim\limits_{x\to6}\left (\frac{f(x)}{3-g(x)}\right )\)
    5. \(\displaystyle \lim\limits_{x\to9}g(f(x))\) \(\lim\limits_{x\to6}f(g(x))\)
    6. \(\displaystyle \lim\limits_{x\to6}g(f(f(x)))\)
    7. \(\displaystyle \lim\limits_{x\to6}f(x)g(x)-f^2(x)+g^2(x)\)

    Using

    \[\begin{align} \lim\limits_{x\to1}f(x)=2 \quad \lim\limits_{x\to10}f(x)=10 \end{align}\]
    \[\begin{align} \lim\limits_{x\to1}g(x)=0 \quad \lim\limits_{x\to10}g(x)=\pi \end{align}\]

    evaluate the limits given in Exercises 14-17, where possible. If it is not possible to know, state so.

    9. \(\displaystyle \lim\limits_{x\to1}f(x)^{g(x)}\)

    10. \(\displaystyle \lim\limits_{x\to10}\cos (g(x))\)

    11. \(\displaystyle \lim\limits_{x\to1}f(x)g(x)\)

    12. \(\displaystyle \lim\limits_{x\to1}g(5f(x))\)

    Answer

    Under construction. 

    Exercise \(\PageIndex{2}\)E:

    Evaluate the following:

    1. \(\displaystyle \lim\limits_{x\to3}x^2-3x+7\)
    2. \(\displaystyle \lim\limits_{x\to\pi}\left ( \frac{x-3}{x+5}\right )^7\)
    3. \(\displaystyle \lim\limits_{x\to\pi /4}\cos x \sin x\)
    4. \(\displaystyle \lim\limits_{x\to 0}\ln x\)
    5. \(\displaystyle \lim\limits_{x\to3}4^{{x^3}-8x}\)
    6. \(\displaystyle \lim\limits_{x\to\pi/6}\csc x\)
    7. \(\displaystyle \lim\limits_{x\to0}\ln (1+x)\)
    8. \(\displaystyle \lim\limits_{x\to\pi}\frac{x^2+3x+5}{5x^2-2x-3}\)
    9. \(\displaystyle \lim\limits_{x\to\pi}\frac{3x+1}{1-x}\)
    10. \(\displaystyle \lim\limits_{x\to6}\frac{x^2-4x-12}{x^2-13x+42}\)
    11. \(\displaystyle \lim\limits_{x\to0}\frac{x^2+2x}{x^2-2x}\)
    12. \(\displaystyle \lim\limits_{x\to2}\frac{x^2+6x-16}{x^2-3x+2}\)
    13. \(\displaystyle \lim\limits_{x\to2}\frac{x^2-5x-14}{x^2+10x+16}\)
    14. \(\displaystyle \lim\limits_{x\to-2}\frac{x^2-5x-14}{x^2+10x+16}\)
    15. \(\displaystyle \lim\limits_{x\to-1}\frac{x^2+9x+8}{x^2-6x-7}\)
    Answer

    Under Construction

    Exercise \(\PageIndex{3}\)E: Limit with indeterminate form

    \( \displaystyle \lim_{x \to 0} \frac{\sqrt{x+4}-2}{x}\)

    Answer

    \( \displaystyle \lim_{x \to 0} \frac{\sqrt{x+4}-2}{x} = \frac{\sqrt{0+4}-2}{0} =\left[\frac{0}{0}\right]\)

    = \( \displaystyle \lim_{x \to 0} \frac{(\sqrt{x+4}-2) (\sqrt{x+4}+2)}{x (\sqrt{x+4}+2)}\)

    = \( \displaystyle \lim_{x \to 0} \frac{((x+4)-4) }{x (\sqrt{x+4}+2)}\)

    = \( \displaystyle \lim_{x \to 0} \frac{x }{x (\sqrt{x+4}+2)}\)

    = \( \displaystyle \lim_{x \to 0} \frac{1 }{(\sqrt{x+4}+2)}= \frac{1 }{(\sqrt{0+4}+2)}= \frac{1 }{4}\).

    Exercise \(\PageIndex{4}\)E:

    Evaluate the following limits:

    1. \(\displaystyle \lim\limits_{x\to4}\frac{1}{|4-x|}\)

    Answer

    \(\displaystyle \infty\)

    2. \(\displaystyle \lim\limits_{x\to-1^-}\sqrt{1-x^2}\)

    Answer

    0

    3. \(\displaystyle \lim\limits_{x\to-1^+}\sqrt{1-x^2}\)

    Answer

    0

    4. \(\displaystyle \lim\limits_{x\to2}\frac{|x-2|}{x^2+x-6}\)

    Answer

    \(\displaystyle \mbox{dne}\)

    5. \(\displaystyle \lim\limits_{x\to2}\frac{\frac{1}{x}-\frac{1}{2}}{x-2}\)

    Answer

    \(\displaystyle \frac{1}{4}\)

    6. \(\displaystyle f(x)=\begin{cases} x^2 & \mbox{if } x \leq 1 \\ 2x & \mbox{if } x > 1\end{cases}\)

    Answer

    Under Construction

    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)

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