Limits Continuity And Differentiability Question 88
Question: If $ f(x)=x+\frac{x}{1+x}+\frac{x}{{{(1+x)}^{2}}}+….to\infty $ , then at $ x=0,f(x) $
Options:
A) Has no limit
B) Is discontinuous.
C) Is continuous but not differentiable
D) Is differentiable
Show Answer
Answer:
Correct Answer: B
Solution:
For $ x\ne 0, $ we have, $ f(x)=x+\frac{x/(1+x)}{1-\frac{1}{1+x}}=x+\frac{x/(1+x)}{x/(1+x)}=x+1 $ For $ x=0,f(x)=0 $ .
Thus, $ f(x)= \begin{cases} x+1, & x\ne 0 \ 0, & x=0 \ \end{cases} . $
Clearly, $ \underset{x\to {0^{-}}}{\mathop{\lim }}f(x)=\underset{x\to {0^{+}}}{\mathop{\lim }}f(x)=1\ne f(0) $ . So, $ f(x) $ is discontinuous and hence not differentiable at $ x=0 $ .
BETA
