Question

$\lim _{x \rightarrow 0} \frac{a^{\sin x}-1}{b^{\sin x}-1}$ equal to

(a) $\frac{a}{b}$

(b) $\frac{b}{a}$

(c) $\frac{\log a}{\log b}$

(d) $\frac{\log b}{\log a}$

Answer 1

$\lim _{x \rightarrow 0} \frac{a^{\sin x}-1}{b^{\sin x}-1}$

$\left(\frac{0}{0} \text { form }\right)$

Using L' Hospital's rule,

$=\lim _{x \rightarrow 0} \frac{a^{\sin x} \log _e a}{b^{\sin x} \log _e b}=\frac{\log _e a}{\log _e b}$

So, The correct option of this question will be (C).