Question

# Consider the following in respect of the function $$f(x) = \left\{\begin{matrix}2+ x, & x \geq 0\\ 2 - x, & x < 0\end{matrix}\right.$$1. $$\displaystyle \lim_{x \rightarrow 1} f(x)$$ 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

Solution

## The correct option is B $$3$$ onlyLeft hand limit$$= \lim_{x\to 1^-}{f(x)}$$$$= \displaystyle \lim_{h\to 0}{f(1-h)}$$$$=\displaystyle \lim_{h\to 0}2+(1-h)=2+1=3$$Right hand limit$$= \displaystyle \lim_{x\to 1^+}{f(x)}$$$$= \displaystyle \lim_{h\to 0}{f(1+h)}$$$$=\displaystyle \lim_{h\to 0}2+(1+h)=2+1=3$$So, the limit exists at x=1Left hand limit$$=\displaystyle \lim_{x\to 0^-}{f(x)}$$$$= \displaystyle \lim_{h\to 0}{f(0-h)}$$$$=\displaystyle \lim_{h\to 0} 2-(0-h)=2-0=2$$Right hand limit$$= \displaystyle \lim_{x\to 0^+}{f(x)}$$$$= \displaystyle \lim_{h\to 0}{f(0+h)}$$$$=\displaystyle \lim_{h\to 0}2+(0+h)=2+0=2$$So, LHL=RHL=f(0)So, f(x) is continuous at x=0Right hand limit$$= \displaystyle \lim_{h\to 0}\dfrac{f(0+h)-f(0)}{h}$$$$= \displaystyle \lim_{h\to 0}\dfrac{(2+(0+h))-2}{h}$$$$=\displaystyle \lim_{h\to 0}\dfrac{h}{h}=1$$Left hand limit$$= \displaystyle \lim_{h\to 0}\dfrac{f(0-h)-f(0)}{-h}$$$$= \displaystyle \lim_{h\to 0}\dfrac{(2-(0-h))-(2)}{-h}$$$$=\displaystyle \lim_{h\to 0}\dfrac{h}{-h}=-1$$LHL$$\neq$$RHLSo, f(x) is not differentiable at x=0The answer is option (B)Mathematics

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