A) a/b
B) b/a
C) \[\frac{\log a}{\log b}\]
D) \[\frac{\log b}{\log a}\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{a}^{\sin x}}-1}{{{b}^{\sin x}}-1},\] Multiplying the numerator and denominator by \[\sin x\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{a}^{x}}-1}{x} \right)\] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{a}^{\sin x}}-1}{{{b}^{\sin x}}-1}\times \frac{\sin x}{\sin x}={{\log }_{e}}a\] \[=\log a\times \frac{1}{{{\log }_{a}}b}=\frac{\log a}{\log b}\]
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