Limits Continuity And Differentiability Question 25
Question: Which of the following is correct for $ f(x)= \begin{matrix} (x-e){2^{-{2^{( \frac{1}{(e-x)} )}},}} & x\ne eatx=e \\ 0, & x=e \\ \end{matrix} . $
Options:
A) f(x) is discontinuous at x = e
B) f(x) is differentiable at x = e
C) f(x) is non-differentiable at x = e
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ f({e^{+}})=\underset{h\to 0}{\mathop{\lim }}(e+h-e){2^{-{2^{\frac{1}{e-(e+h)}}}}} $
$ =\underset{h\to 0}{\mathop{\lim }}(h){2^{-{2^{-\frac{1}{h}}}}}=0\times 1=0 $
$ ( As for h\to 0,-\frac{1}{h}\to -\infty \Rightarrow {2^{-\frac{1}{h}}}\to 0 ) $
$ f({e^{-}})=\underset{h\to 0}{\mathop{\lim }}(-h){2^{-{2^{\frac{1}{h}}}}}=0\times 0=0 $
Hence, $ f(x) $ is continuous at $ x=e $ .
$ f’({e^{+}})=\underset{h\to 0}{\mathop{\lim }}\frac{f(e+h)-f(e)}{h}=\underset{h\to 0}{\mathop{\lim }}\frac{h\times {2^{-{2^{\frac{1}{h}}}}}-0}{h} $
$ =\underset{h\to 0}{\mathop{\lim }}{2^{-{2^{\frac{1}{h}}}}}=1 $
$ f’({e^{-}})=\underset{h\to 0}{\mathop{\lim }}\frac{f(e-h)-f(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}\frac{(-h){2^{-{2^{\frac{1}{h}}}}}-0}{-h} $
$ =\underset{h\to 0}{\mathop{\lim }}{2^{-{2^{\frac{1}{h}}}}}=0 $ .
Hence, $ f(x) $ is non-differentiable at $ x=e $ .
BETA
