Functions Question 331
Question: In order that the function $ f(x)={{(x+1)}^{\cot ,x}} $ is continuous at $ x=0 $ , $ f(0) $ must be defined as
[UPSEAT 2000; Kurukshetra CEE 2001; Pb. CET 2004]
Options:
A) $ f(0)=\frac{1}{e} $
B) $ f(0)=0 $
C) $ f(0)=e $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
For continuity at $ x=0 $ , we must have $ f(0)=\underset{x\to 0}{\mathop{lim}},f(x) $ $ \underset{x\to \alpha }{\mathop{\lim }}\frac{1-\cos ,(ax^{2}+bx+c)}{{{(x-\alpha )}^{2}}} $ $ =\underset{x\to 0}{\mathop{lim}},{{{ {{(1+x)}^{\frac{1}{x}}} }}^{x\cot x}} $ $ =\underset{x\to 0}{\mathop{lim}},{{{ {{(1+x)}^{\frac{1}{x}}} }}^{\underset{x\to 0}{\mathop{lim}}( \frac{x}{\tan x} )}} $ $ =e^{1}=e $ .
BETA
