Differentiation Question 184
Question: $ \underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}{{[1+{{(cosx)}^{\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)logcosx $
$ \log (y)=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\frac{\log (cosx)}{\sec (x)}\left( \frac{\infty }{\infty } \text{ form} \right) $
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
