Q. 2 - Calculus

Question

Construct Examples

Step-by-Step Solution

Answer

 (a) a function that approaches infinity is $\underset{x\to 3}{\lim }\frac{2x}{x-3}.$

(b) Examples of limits that can be solved by using special trigonometric limits are $\underset{x\to 0}{\underset{x\to 0}{\lim 4}\left(\frac{\sin \left(3x\right)}{x}\right) \& \lim 4}\left(\frac{1-\cos \left(3x\right)}{x}\right).$

(c) functions that are discontinuous at $x=3$are the following.

removable discontinuity: $f\left(x\right)=\frac{x^2-9}{x-3}.$

jump discontinuity:$f\left(x\right)=\left\{\begin{array}{ll}x+2& x<3\\ x+5& x>3\end{array}\right.$

infinite discontinuity:$f\left(x\right)=\frac{1}{x-3}.$

Step 1. Given information.

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

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

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

Step 2. Part (a)

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

find the left-hand and right-hand limit.

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

Step 3. Part (b)

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

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

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

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

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

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

Step 4. Part (c)

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 $f\left(x\right)=\frac{1}{x-3}.$

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