mathop {lim }limits_{x to 0} frac{{{a^{sin x} | vedclass

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Question

(C) We know that $\mathop {\lim }\limits_{u 	o 0} \frac{k^u - 1}{u} = \ln k$.Given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{a^{\sin x} - 1}

Vedclass · Accepted Answer

$\frac{\log a}{\log b}$

Vedclass · Answer

(C) We know that $\mathop {\lim }\limits_{u 	o 0} \frac{k^u - 1}{u} = \ln k$.Given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{a^{\sin x} - 1}{b^{\sin x} - 1}$.Multiply and divide by $\sin x$: $\mathop {\lim }\limits_{x 	o 0} \left( \frac{a^{\sin x} - 1}{\sin x} 	imes \frac{\sin x}{b^{\sin x} - 1} ight)$.As $x 	o 0$,$\sin x 	o 0$. Thus,the expression becomes $\ln a 	imes \frac{1}{\ln b}$.Therefore,the result is $\frac{\log a}{\log b}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{a^{\sin x}} - 1}}{{{b^{\sin x}} - 1}} = $

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