Differentiation Question 493
Question: $ \underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }},{{[1+{{(cos,x)}^{\cos x}}]}^{2}} $ is equal to
Options:
A) Does not exist
1
e
4
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Given $ \underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }},{{[1+{{(cosx)}^{\cos x}}]}^{2}} $ Let $ y=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }},{{(cosx)}^{\cos x}} $ $ \log (y)=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }},(cosx)\log,cos,x $ $ \log (y)=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }},\frac{\log (cosx)}{\sec (x)}( \frac{0}{0}form ) $ Applying L?Hospital?s rule $ \log (y)=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }},\frac{-\sin x}{\cos x(sec,x,tan,x)} $ $ =\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }},(-cosx)=0. $ $ \therefore y=e^{0}=1 $ Now, limits is $ {{(1+1)}^{2}}=2^{2}=4 $
BETA
