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# 1.3E Exercises

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### Exercise $$\PageIndex{1}$$

Using:

\begin{align}\lim\limits_{x\to9}f(x)=6 \qquad \lim\limits_{x\to6}f(x)=9 \\ \lim\limits_{x\to9}g(x)=3 \qquad \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.  $$\lim\limits_{x\to9}(f(x)+g(x))$$
2.  $$\lim\limits_{x\to9}(3f(x)/g(x))$$
3.  $$\lim\limits_{x\to9} \left ( \frac{f(x)-2g(x)}{g(x)}\right )$$
4. $$\lim\limits_{x\to6}\left (\frac{f(x)}{3-g(x)}\right )$$
5. $$\lim\limits_{x\to9}g(f(x))$$ $$\lim\limits_{x\to6}f(g(x))$$
6.  $$\lim\limits_{x\to6}g(f(f(x)))$$
7.  $$\lim\limits_{x\to6}f(x)g(x)-f^2(x)+g^2(x)$$

Using

\begin{align}\lim\limits_{x\to1}f(x)=2 \qquad \lim\limits_{x\to10}f(x)=1 \\ \lim\limits_{x\to1}g(x)=0 \qquad \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. $$\lim\limits_{x\to1}f(x)^{g(x)}$$

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

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

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

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### Exercise $$\PageIndex{2}$$

Evaluate the following:

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

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### Exercise $$\PageIndex{3}$$: Limit with indeterminate form

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

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

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

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

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

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

## 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)