Differentiation Question 500
Question: The limit $ \underset{x\to 0}{\mathop{\lim }},{{(cosx)}^{1/\sin x\frac{1}{\sin x}}} $ is equal to
Options:
A) $ e $
B) $ {e^{-1}} $
C) 1
D) Does not exist
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \underset{x\to 0}{\mathop{\lim }},{{(cos)}^{\frac{1}{\sin x}}} $ is $ {1^{\infty }} $ form $ ={e^{\underset{x\to \infty }{\mathop{Lim}},(cosx-1)\frac{1}{\sin x}}} $ $ ={e^{\underset{x\to 0}{\mathop{Lim}},\frac{-2{{\sin }^{2}}x/2}{2{{\sin }^{x}}/{2^{\cos x}}/2}}}={e^{\underset{x\to 0}{\mathop{Lim}},( -\tan \frac{x}{2} )}}=e^{o}=1 $
BETA
