Differentiation Question 500
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}} $
1
D) Does not exist
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{x\to 0}{\mathop{\lim }},{{(cos)}^{\frac{1}{\sin x}}} $ is $ {1^{\infty }} $ form $ ={e^{\underset{x\to 0}{\mathop{Lim}},(cosx-1)\frac{1}{\sin x}}} $ $ ={e^{\underset{x\to 0}{\mathop{Lim}},\frac{-2{{\sin }^{2}}\frac{x}{2}}{2{{\sin }^{2}}\frac{x}{2}}}={e^{\underset{x\to 0}{\mathop{Lim}},( -\tan \frac{x}{2} )}}=e^{0}=1 $
BETA
