A) Does not exist
B) 1
C) e
D) 4
Correct Answer: D
Solution :
[d] Given \[\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,{{[1+{{(cosx)}^{\cos x}}]}^{2}}\] Let \[y=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,{{(cosx)}^{\cos x}}\] \[\log (y)=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,(cosx)log\,cos\,x\] \[\log (y)=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\frac{\log (cosx)}{\sec (x)}\left( \frac{\infty }{\infty }form \right)\] 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 \[{{(1+1)}^{2}}={{2}^{2}}=4.\]
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