\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{a}^{\sin x}}-1}{{{b}^{\sin x}}-1}=\] [Karnataka CET 2000]

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{a}^{\sin x}}-1}{{{b}^{\si

1.	\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{a}^{\sin x}}-1}{{{b}^{\sin x}}-1}=\] [Karnataka CET 2000]
A.	\[\frac{a}{b}\]
B.	\[\frac{b}{a}\]
C.	\[\frac{\log a}{\log b}\]
D.	\[\frac{\log b}{\log a}\]
Answer» D. \[\frac{\log b}{\log a}\]

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\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{a}^{\sin x}}-1}{{{b}^{\sin x}}-1}=\] [Karnataka CET 2000]

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