lim _{x rightarrow 0} frac{3^{sin x}-2^{tan x | vedclass

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Question

(D) We use the standard limit $\lim _{u \rightarrow 0} \frac{a^u - 1}{u} = \ln a$. The given expression is $\lim _{x \rightarrow 0} \frac{3^{\sin x}-2

Vedclass · Accepted Answer

$\log _e \frac{3}{2}$

Vedclass · Answer

(D) We use the standard limit $\lim _{u ightarrow 0} \frac{a^u - 1}{u} = \ln a$. The given expression is $\lim _{x ightarrow 0} \frac{3^{\sin x}-2^{	an x}}{\sin x}$. Dividing the numerator and denominator by $x$,we get: $\lim _{x ightarrow 0} \frac{\frac{3^{\sin x}-1}{x} - \frac{2^{	an x}-1}{x}}{\frac{\sin x}{x}}$. Since $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x ightarrow 0} \frac{	an x}{x} = 1$,we have: $\lim _{x}$ ${ightarrow 0} \left( \frac{3^{\sin x}-1}{\sin x} \cdot \frac{\sin x}{x} - \frac{2^{	an x}-1}{	an x} \cdot \frac{	an x}{x} ight) / \frac{\sin x}{x}$. This simplifies to $\ln 3 - \ln 2 = \ln \left( \frac{3}{2} ight)$.

$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}=$

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