A) \[\frac{a}{b}\]
B) \[\frac{b}{a}\]
C) \[\frac{\log a}{\log b}\]
D) \[\frac{\log b}{\log a}\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\text{lim}}}\,\frac{{{a}^{\sin x}}-1}{{{b}^{\sin x}}-1}=\underset{x\to 0}{\mathop{\text{lim}}}\,\frac{{{a}^{\sin x}}-1}{\sin x}\times \frac{\sin x}{{{b}^{\sin x}}-1}\] \[\] \[={{\log }_{e}}a\times \frac{1}{{{\log }_{e}}b}=\frac{\log a}{\log b}\].
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