$\lim _{x \rightarrow 0} \frac{(2+3 \sin x)^x-2^x}{\tan ^2 x-4 x^3}$.
【答案】 【参考解答】: 由 $\tan ^2 x-4 x^3 \sim \tan ^2 x \sim x^2(x \rightarrow 0)$, 又
$$
\mathrm{e}^x-1 \sim x \sim \ln (1+x)(x \rightarrow 0)
$$
故改写极限式,得
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{(2+3 \sin x)^x-2^x}{\tan ^2 x-4 x^3}=\lim _{x \rightarrow 0} \frac{2^x\left[\left(1+\frac{3}{2} \sin x\right)^x-1\right]}{x^2} \\
=& \lim _{x \rightarrow 0} 2^x \cdot \lim _{x \rightarrow 0} \frac{\left.e^{x \ln \left(1+\frac{3}{2} \sin x\right)}-1\right]}{x^2}=\lim _{x \rightarrow 0} \frac{x \ln \left(1+\frac{3}{2} \sin x\right)}{x^2} \\
=& \lim _{x \rightarrow 0} \frac{\frac{3}{2} \sin x}{x}=\frac{3}{2}
\end{aligned}
$$


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