The Limit Laws

In the following exercises (1-2), use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

1. Use constant multiple law and difference law: $\underset{x\to 0}{\lim}(4x^2-2x+3)=4\underset{x\to 0}{\lim}x^2-2\underset{x\to 0}{\lim}x+\underset{x\to 0}{\lim}3=3$
2. Use root law: $\underset{x\to -2}{\lim}\sqrt{x^2-6x+3}=\sqrt{\underset{x\to -2}{\lim}(x^2-6x+3)}=\sqrt{19}$
3. $49$
4. $1$
5. $-\frac{5}{7}$
6. $\underset{x\to 4}{\lim}\frac{x^2-16}{x-4}=\frac{16-16}{4-4}=\frac{0}{0}$; then, $\underset{x\to 4}{\lim}\frac{x^2-16}{x-4}=\underset{x\to 4}{\lim}\frac{(x+4)(x-4)}{x-4}=8$
7. $\underset{x\to 6}{\lim}\frac{3x-18}{2x-12}=\frac{18-18}{12-12}=\frac{0}{0}$; then, $\underset{x\to 6}{\lim}\frac{3x-18}{2x-12}=\underset{x\to 6}{\lim}\frac{3(x-6)}{2(x-6)}=\frac{3}{2}$
8. $\underset{t \to 9}{\lim}\frac{t-9}{\sqrt{t}-3}=\frac{9-9}{3-3}=\frac{0}{0}$; then, $\underset{t\to 9}{\lim}\frac{t-9}{\sqrt{t}-3}=\underset{t\to 9}{\lim}\frac{t-9}{\sqrt{t}-3}\frac{\sqrt{t}+3}{\sqrt{t}+3}=\underset{t\to 9}{\lim}(\sqrt{t}+3)=6$
9. $\underset{\theta \to \pi}{\lim}\frac{\sin \theta}{\tan \theta}=\frac{\sin \pi}{\tan \pi}=\frac{0}{0}$; then, $\underset{\theta \to \pi}{\lim}\frac{\sin \theta}{\tan \theta}=\underset{\theta \to \pi}{\lim}\frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}}=\underset{\theta \to \pi}{\lim}\cos \theta =-1$.
10. $\underset{x\to 1/2}{\lim}\frac{2x^2+3x-2}{2x-1}=\frac{\frac{1}{2}+\frac{3}{2}-2}{1-1}=\frac{0}{0}$; then, $\underset{x\to 1/2}{\lim}\frac{2x^2+3x-2}{2x-1}=\underset{x\to 1/2}{\lim}\frac{(2x-1)(x+2)}{2x-1}=\frac{5}{2}$
11. $\underset{x\to 6}{\lim}2f(x)g(x)=2\underset{x\to 6}{\lim}f(x)\underset{x\to 6}{\lim}g(x)=72$
12. $\underset{x\to 6}{\lim}(f(x)+\frac{1}{3}g(x))=\underset{x\to 6}{\lim}f(x)+\frac{1}{3}\underset{x\to 6}{\lim}g(x)=7$
13. $\underset{x\to 6}{\lim}\sqrt{g(x)-f(x)}=\sqrt{\underset{x\to 6}{\lim}g(x)-\underset{x\to 6}{\lim}f(x)}=\sqrt{5}$
14. $\underset{x\to 6}{\lim}[(x+1)\cdot f(x)]=(\underset{x\to 6}{\lim}(x+1))(\underset{x\to 6}{\lim}f(x))=28$.
15. 
    1. $9$
    2. $7$
16. 
    1. $1$
    2. $1$
17. $\underset{x\to -3^-}{\lim}(f(x)-3g(x))=\underset{x\to -3^-}{\lim}f(x)-3\underset{x\to -3^-}{\lim}g(x)=0+6=6$
18. $\underset{x\to -5}{\lim}\frac{2+g(x)}{f(x)}=\frac{2+(\underset{x\to -5}{\lim}g(x))}{\underset{x\to -5}{\lim}f(x)}=\frac{2+0}{2}=1$
19. $\underset{x\to 1}{\lim}\sqrt[3]{f(x)-g(x)}=\sqrt[3]{\underset{x\to 1}{\lim}f(x)-\underset{x\to 1}{\lim}g(x)}=\sqrt[3]{2+5}=\sqrt[3]{7}$
20. $\underset{x\to -9}{\lim}(x\cdot f(x)+2g(x))=(\underset{x\to -9}{\lim}x)(\underset{x\to -9}{\lim}f(x))+2\underset{x\to -9}{\lim}(g(x))=(-9)(6)+2(4)=-46$

Continuity

For the following exercises (1-4), determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

1. The function is defined for all $x$ in the interval $(0,\infty)$.
2. Removable discontinuity at $x=0$; infinite discontinuity at $x=1$
3. $f(x)=\dfrac{5}{e^x-2}$ Infinite discontinuity at $x=\ln 2$
4. $H(x)= \tan 2x$ Infinite discontinuities at $x=\frac{(2k+1)\pi}{4}$, for $k=0, \, \pm 1, \, \pm 2, \, \pm 3, \cdots$
5. No. It is a removable discontinuity.
6. Yes. It is continuous.
7. Yes. It is continuous.
8. $k=-5$
9. $k=-1$
10. $k=\frac{16}{3}$
11. 

    It is not possible to redefine $f(1)$ since the discontinuity is a jump discontinuity.

12. Answers may vary; see the following example:

    

13. Answers may vary; see the following example:

    /li>

14. False. It is continuous over $(−\infty,0) \cup (0,\infty)$.
15. False. Consider $f(x)=\begin{cases} x & \text{ if } \, x \ne 0 \\ 4 & \text{ if } \, x = 0 \end{cases}$
16. False. Consider $f(x)= \cos (x)$ on $[-\pi, 2\pi]$.
17. False. The IVT does not work in reverse! Consider $(x-1)^2$ over the interval $[-2,2]$.
18. For all values of $a, \, f(a)$ is defined, $\underset{\theta \to a}{\lim}f(\theta)$ exists, and $\underset{\theta \to a}{\lim}f(\theta)=f(a)$. Therefore, $f(\theta)$ is continuous everywhere.
19. Nowhere

The Precise Definition of a Limit

In the following exercises (1-2), write the appropriate $\varepsilon$–$\delta$ definition for each of the given statements.

1. For every $\varepsilon > 0$, there exists a $\delta > 0$ so that if $0 < |t-b| < \delta$, then $|g(t)-M| < \varepsilon$
2. For every $\varepsilon > 0$, there exists a $\delta > 0$ so that if $0 < |x-a| < \delta$, then $|\phi(x)-A| < \varepsilon$
3. $\delta \le 0.25$
4. $\delta \le 2$
5. $\delta \le 1$
6. $\delta < 0.3900$
7. Let $\delta =\varepsilon$. If $0 < |x-3| < \varepsilon$, then $|x+3-6|=|x-3| < \varepsilon$.
8. Let $\delta =\sqrt[4]{\varepsilon}$. If $0 < |x| < \sqrt[4]{\varepsilon}$, then $|x^4|=x^4 < \varepsilon$.
9. Let $\delta =\varepsilon^2$. If $5-\varepsilon^2 < x < 5$, then $|\sqrt{5-x}|=\sqrt{5-x} < \varepsilon$.
10. Let $\delta =\varepsilon/5$. If $1-\varepsilon/5 < x < 1$, then $|f(x)-3|=5x-5 < \varepsilon$.
11. Let $\delta =\sqrt{\frac{3}{N}}$. If $0 < |x+1| < \sqrt{\frac{3}{N}}$, then $f(x)=\frac{3}{(x+1)^2} > N$.
12. $f(x)-g(x)=f(x)+(-1)g(x)$
13. Answers may vary.