A) \[\frac{1}{2}\]
B) \[\frac{1}{e}\]
C) \[e\]
D) 1
Correct Answer: C
Solution :
Let\[{{\ell }_{1}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{a{{(}^{\sin x}}-1)\sin x}{(1+x-\cos x)\sin x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{({{a}^{\sin x}}-1)}{\sin x(1+x-\cos x)}\cdot \frac{(\sin x)}{x}.x\] \[=(\ell na)\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{(1+x-\cos x)}\] \[=(\ell na)\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\frac{(1-\cos x)}{x}+\frac{x}{x}}=\ell na\] \[=(\ell na)\underset{x\to 0}{\mathop{\lim }}\,\frac{x-1}{\ell n(x)}\] Put\[x=1+h\] \[{{\ell }_{2}}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{\ell n(1+h)}=1\] Hence \[\ell na=1\Rightarrow a=e\].
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