Question

The function $f\left(x\right)=\left|x\right|+\frac{|x|}{x}$ is:

• A. discountinuous at origin because |x| is discontinuous there.
• B. continuous at origin.
• C. discontinuous at origin because both $|x|$ and $\frac{|x|}{x}$ are discontinuous there.
• D. discontinuous at the origin because $\frac{|x|}{x}$ discontinuous there.

Answer 1

Answer: (D)

Let $f\left(x\right)=\left|x\right|+\frac{|x|}{x}$
Let us consider
$f(x)=f_1(x)+f_2(x)$
Where $f_1(x)=|x|$ and $f_2(x)=\frac{|x|}{x}$
Now at x = 0
$f_1(0)=0$
LHL = $\underset{x\to 0^{-}}{\lim }|x|=0$ and
RHL $=\underset{x\to 0^{+}}{\lim }|x|=0$
$\therefore f_1(x)$ is continuous at $x=0$
Now consider $f_2(x)=\frac{|x|}{x}$
Now LHL = $\underset{x\to 0^{-}}{\lim }\frac{|x|}{x}$
Put $x=0-h$
$\underset{x\to 0^{-}}{\lim }\frac{|0-h|}{0-h}=\underset{x\to 0^{-}}{\lim }-\frac{h}{h}=-1$
Now RHL = $\underset{x\to 0^{+}}{\lim }\frac{|x|}{x}$
$=\underset{x\to 0^{+}}{\lim }\frac{|0+h|}{0+h}$ [By putting x = 0 + h]
$=\underset{x\to 0}{\lim }\frac{h}{h}=1$
Since LHL $\ne$ RHL
$\therefore f_2(x)$ is discontinuous at origin.
$\therefore f\left(x\right)=\left|x\right|+\frac{|x|}{x}$ is discontinuous at origin because $\frac{|x|}{x}$ is discontinuous.