Limits Continuity And Differentiability Question 28
Question: Consider the following in respect of the function $ f(x)= \begin{cases} 2+x, & x\ge 0 \\ 2-x, & x<0 \\ \end{cases} . $
- $ \underset{x\to 1}{\mathop{\lim f}}(x) $ does not exist.
- f(x) is differentiable at x = 0
- f(x) is continuous at x = 0 Which of the above statements is/are correct?
Options:
A) 1 only
B) 3 only
C) 2 and 3 only
D) 1 and 3 only
Show Answer
Answer:
Correct Answer: D
Solution:
For $ x\ge 0 $
$ \underset{x\to 1}{\mathop{\lim }}f(x)=\underset{x\to 1}{\mathop{\lim }}2+x=2+1=3 $ For $ x<0 $
$ \underset{x\to 1}{\mathop{\lim }}f(x)=\underset{x\to 1}{\mathop{\lim }}2-x=2-1=1 $ So, $ \underset{x\to 1}{\mathop{\lim }}f(x) $ does not exist. At $ x=0 $
$ RHL:\underset{h\to {0^{+}}}{\mathop{\lim }}f(0+h)=\underset{h\to 0}{\mathop{\lim }}2+h=2 $
$ LHL:\underset{h\to {0^{-}}}{\mathop{\lim }}f(0-h)=\underset{h\to 0}{\mathop{\lim }}2-h=2 $
$ f(0)=2+0=2. $ So, RHL = LHL = f(0)
$ \Rightarrow f(x) $ is continuous at $ x=0 $ Differentiability at $ x=0 $
$ LHD:\underset{h\to {0^{-}}}{\mathop{\lim }}\frac{f(0-h)-f(0)}{-h}=\underset{h\to {0^{-}}}{\mathop{\lim }}\frac{2+h-2}{-h} $
$ =\frac{-h}{h}=-1 $
$ RHD:\underset{h\to {0^{+}}}{\mathop{\lim }}\frac{f(0+h)-f(0)}{h}=\underset{h\to {0^{+}}}{\mathop{\lim }}\frac{2+h-2}{h}=1 $
Since $ LHD\ne RHD $
So, $ f(x) $ is not differentiable at $ x=0 $
BETA
