Question

Consider the following in respect of the function $f\left(x\right)=\{\begin{array}{cc}2+x,& x\ge 0\\ 2-x,& x<0\end{array}$
1. $\underset{x\to 1}{lim}f\left(x\right)$ does not exist.
2. f(x) is differentiable at x = 0.
3. f(x) is continuous at x = 0.
Which of the above statements is/are correct?

• A. $1$ only
• B. $3$ only
• C. $2$ and $3$ only
• D. $1$ and $3$ only

Answer 1

The correct option is B $3$ only

Left hand limit$={lim}_{x\to 1^{-}}f\left(x\right)$

$=\underset{h\to 0}{lim}f(1-h)$

$=\underset{h\to 0}{lim}2+(1-h)=2+1=3$

Right hand limit$=\underset{x\to 1^{+}}{lim}f\left(x\right)$

$=\underset{h\to 0}{lim}f(1+h)$

$=\underset{h\to 0}{lim}2+(1+h)=2+1=3$

So, the limit exists at x=1

Left hand limit$=\underset{x\to 0^{-}}{lim}f\left(x\right)$

$=\underset{h\to 0}{lim}f(0-h)$

$=\underset{h\to 0}{lim}2-(0-h)=2-0=2$

Right hand limit$=\underset{x\to 0^{+}}{lim}f\left(x\right)$

$=\underset{h\to 0}{lim}f(0+h)$

$=\underset{h\to 0}{lim}2+(0+h)=2+0=2$

So, LHL=RHL=f(0)

So, f(x) is continuous at x=0

Right hand limit$=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$

$=\underset{h\to 0}{lim}\frac{(2+(0+h\left)\right)-2}{h}$

$=\underset{h\to 0}{lim}\frac{h}{h}=1$

Left hand limit$=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{(2-(0-h\left)\right)-\left(2\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{h}{-h}=-1$

LHL$\ne$RHL

So, f(x) is not differentiable at x=0

The answer is option (B)