Differentiation Question 196
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
