Functions Question 602
Question: If $ f(x)= \begin{cases} & {2^{1/x}},for,x\ne 0 \\ & ,3,for,x=0 \\ \end{cases} . $ , then
Options:
A) $ \underset{x\to 0+}{\mathop{\lim }},f(x)=0 $
B) $ \underset{x\to 0-}{\mathop{\lim }},f(x)=\infty $
C) $ f(x) $ is continuous at $ x=0 $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{x\to 0+}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }},{2^{1/h}}=\infty $ $ \underset{x\to 0-}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }},{2^{-1/h}}=\underset{h\to 0}{\mathop{\lim }}\frac{1}{{2^{1/h}}}=0 $ .
BETA
