Functions Question 228
Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{{a^{\sin x}}-1}{{b^{\sin x}}-1}= $
[Karnataka CET 2000]
Options:
A) $ \frac{a}{b} $
B) $ \frac{b}{a} $
C) $ \frac{\log a}{\log b} $
D) $ \frac{\log b}{\log a} $
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Answer:
Correct Answer: C
Solution:
$ \underset{x\to 0}{\mathop{lim}},\frac{{a^{\sin x}}-1}{{b^{\sin x}}-1}=\underset{x\to 0}{\mathop{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} $ .
BETA
