Functions Question 561
Question: The value of f at $ x=0 $ so that the function $ f(x)=\frac{2^{x}-{2^{-x}}}{x},x\ne 0 $ , is continuous at $ x=0 $ , is
[Kerala (Engg.) 2005]
Options:
A) 0
B) log 2
C) 4
D) $ e^{4} $
E) log 4
Show Answer
Answer:
Correct Answer: E
Solution:
$ f(0)=\underset{x\to 0}{\mathop{\lim }},f(x)=\underset{x\to 0}{\mathop{\lim }},( \frac{2^{x}-{2^{-x}}}{x} )=\underset{x\to 0}{\mathop{\lim }},[ \frac{(2^{x}+{2^{-x}}){\log_{e}}2}{1} ] $ $ =(2^{0}+2^{0}){\log_{e}}2 $ $ =(1+1){\log_{e}}2 $ $ =2{\log_{e}}2={\log_{e}}4 $ .
BETA
