Question

Function $f\left(x\right)=\{\begin{array}{c}\frac{x+\left|x\right|}{x}\text{; for}x\ne 0\\ 3\text{; for}x=0\end{array}$

• A. It is continuous at $x=0$
• B. Its $\underset{x\to 0}{lim}f\left(x\right)\ne f\left(0\right)$
• C. It is not defined at $x=0$
• D. It is discontinuous at $x=0$

Answer 1

The correct option is D It is discontinuous at $x=0$

$f\left(x\right)=\frac{x+|x|}{x}$ for $x\ne 0$

$=3$ for $x=0$

For $x\to 0$ and $x>0,|x|=x$ and thus,

$f\left(x\right)=\frac{x+x}{x}=2$

For $x\to 0$ and $x<0,|x|=-x$ and so,

$f\left(x\right)=\frac{x-x}{x}=0$

Since both the values are not the same and also not equal to $3,f\left(x\right)$ is dicontinuous at $x=0.$