A) \[e\]
B) \[{{e}^{-1}}\]
C) 1
D) Does not exist
Correct Answer: C
Solution :
[c] \[\underset{x\to 0}{\mathop{\lim }}\,{{(cos)}^{\frac{1}{\sin x}}}\] is \[{{1}^{\infty }}\] form \[={{e}^{\underset{x\to \infty }{\mathop{Lim}}\,(cosx-1)\frac{1}{\sin x}}}\] \[={{e}^{\underset{x\to 0}{\mathop{Lim}}\,\frac{-2{{\sin }^{2}}x/2}{2{{\sin }^{x}}/{{2}^{\cos x}}/2}}}={{e}^{\underset{x\to 0}{\mathop{Lim}}\,\left( -\tan \frac{x}{2} \right)}}={{e}^{o}}=1\]
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