lim _{x rightarrow 0} frac{a^x-1}{sin (x)} = | vedclass

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Question

(A) We are given the limit $\lim _{x \rightarrow 0} \frac{a^x-1}{\sin x}$.Dividing both the numerator and the denominator by $x$,we get:$\lim _{x \rig

Vedclass · Accepted Answer

$\log _e (a)$

Vedclass · Answer

(A) We are given the limit $\lim _{x \rightarrow 0} \frac{a^x-1}{\sin x}$.Dividing both the numerator and the denominator by $x$,we get:$\lim _{x \rightarrow 0} \frac{\frac{a^x-1}{x}}{\frac{\sin x}{x}}$.Using the standard limits $\lim _{x \rightarrow 0} \frac{a^x-1}{x} = \log _e a$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have:$\frac{\log _e a}{1} = \log _e a$.Thus,the correct option is $A$.

$\lim _{x \rightarrow 0} \frac{a^x-1}{\sin (x)} = $

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