Question

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

(A) $\log \left(\frac{a}{b}\right)$

(B) $\log \left(\frac{b}{a}\right)$

(C) $\log (a b)$

(D) $\log (a+b)$

Answer 1

$\lim _{x \rightarrow 0} \frac{a^x-b^x}{e^x-1} =\lim _{x \rightarrow 0} \frac{a^x-b^x}{x} \cdot \frac{x}{e^x-1}$

$=\lim _{x \rightarrow 0}\left[\frac{a^x-1}{x}-\frac{b^x-1}{x}\right] \frac{x}{e^x-1}$

$=\left[\lim _{x \rightarrow 0}\left(\frac{a^x-1}{x}\right)-\lim _{x \rightarrow 0}\left(\frac{b^x-1}{x}\right)\right] \lim _{x \rightarrow 0} \frac{1}{\frac{e^x-1}{x}}$

$=\left(\log _e a-\log _e b\right) \frac{1}{1}$

$ =\log _e\left(\frac{a}{b}\right)$

So, The correct option will be (A).