Limits Continuity And Differentiability Question 30
Question: If $ f(x)={{(x+1)}^{\cot x}} $ is continuous at $ x=0 $ , then what is f (0) equal to?
Options:
A) 1
B) e
C) $ \frac{1}{e} $
D) $ e^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
For a function to be continuous at a point the limit should exist and should be equal to the value of the function at that point. Here point is x = 0 and $ \underset{x\to 0}{\mathop{\lim }}f(x)=\underset{x\to 0}{\mathop{\lim }}{{(x+1)}^{\cot x}} $
$ =\underset{x\to 0}{\mathop{\lim }}{{(x+1)}^{\cot x}}=\underset{x\to 0}{\mathop{\lim }}{{(1+x)}^{\frac{1}{x}.x\cot x}} $
$ =\underset{x\to 0}{\mathop{\lim }}{{(x+1)}^{\frac{1}{x}\underset{x\to 0}{\mathop{\lim }}\frac{x}{\tan x}}}=e^{1}=e $ Since limiting value of $ f(x)=e $ , when $ x\to 0,f(0) $ should also be equal to e.
BETA
