91Ó°ÊÓ

(a) Limits approaches $∞\mathrm{as}x→c^{+}\mathrm{and}-∞\mathrm{as}x→c^{-}.$

(b) Limits can be solved by using the special trigonometric limits from Theorem 1.35.

(c) the function is discontinuous at $x=3.$

Consider a limit $\underset{x→3}{\lim }\frac{2x}{x-3}.$

find the left-hand and right-hand limit.

$$\begin{gathered} \underset{x→3^{+}}{\lim }\frac{2x}{x-3}=∞ \\ \underset{x→3^{-}}{\lim }\frac{2x}{x-3}=-∞ \end{gathered}$$

Consider a limit $\underset{x→0}{\lim 4}\left(\frac{\sin \left(3x\right)}{x}\right).$

Solve the limit using special trigonometric limits $\underset{x→0}{\lim }\frac{\sin x}{x}=1$.

$$\begin{gathered} \underset{x→0}{\lim 4}\left(\frac{\sin \left(3x\right)}{x}\right)=\underset{x→0}{\lim }\left(\frac{\sin \left(3x\right)}{3x}\right)\left(12\right) \\ =1\left(12\right) \\ =12 \end{gathered}$$

Consider a limit $\underset{x→0}{\lim 4}\left(\frac{1-\cos \left(3x\right)}{x}\right)$

Solve the limit using special trigonometric limits $\underset{x→0}{\lim }\frac{1-coxx}{x}=1$

$$\begin{gathered} \underset{x→0}{\lim 4}\left(\frac{1-\cos \left(3x\right)}{x}\right)=\underset{x→0}{\lim }\left(\frac{1-\cos \left(3x\right)}{3x}\right)\left(12\right) \\ =1\left(12\right) \\ =12 \end{gathered}$$

Consider a function $f\left(x\right)=\frac{x^2-9}{x-3}.$

The function has a removable discontinuity at $x=3.$

Consider a function $f\left(x\right)=\left\{\begin{array}{ll}x+2& x<3\\ x+5& x>3\end{array}\right.$

The function has a jump discontinuity at $x=3.$

Consider a function localid="1648719058593" $f\left(x\right)=\frac{1}{x-3}.$

The function has a vertical asymptote at $x=3,$so the function is an infinite discontinuity at $x=3.$