Chapter 14.3, Problem 60AYU

To determine

To find: Whether $f$ is continuous at $c$.

Answer to Problem 60AYU

Continuous.

Explanation of Solution

Given:

$f\left(x\right)=3\cos x;x<0$

$f\left(x\right)=3;x=\text{ 0}$

$f\left(x\right)=\frac{x^3+3x^2}{x^2};x>\text{ 0}$

$c=0$

Calculation:

We see that,

$\underset{x\to c^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\left(3\cos x\right)=\left(3\cos 0\right)=3\left(1\right)=3$

Also,

$\underset{x\to c^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\left(\frac{x^3+3x^2}{x^2}\right)=\underset{x\to 0^{+}}{\lim }\frac{x^2\left(x+3\right)}{x^2}=\underset{x\to 0^{+}}{\text{lim}}\left(x+3\right)=0+3=3$

Again,

$f\left(c\right)=f\left(0\right)=3$; by definition.

Since $\underset{x\to c^{-}}{\lim }f\left(x\right)=\underset{x\to c^{+}}{\lim }f\left(x\right)=f\left(c\right)$, therefore $f$ is continuous at $c=0$.