$\text{A.}$ $\lim _{h \rightarrow 0} \frac{1}{h^{2}} f(1-\cos h)$ 存在. $\text{B.}$ $\lim _{h \rightarrow 0} \frac{1}{h} f\left(1-\mathrm{e}^{h}\right)$ 存在. $\text{C.}$ $\lim _{h \rightarrow 0} \frac{1}{h^{2}} f(h-\sin h)$ 存在. $\text{D.}$ $\lim _{h \rightarrow 0} \frac{1}{h}[f(2 h)-f(h)]$ 存在.

B

#### 解析：

$\lim _{h \rightarrow 0} \frac{1}{h} f\left(1-e^{h}\right)=e^{h}=x \lim _{x \rightarrow 0} \frac{f(x)}{\ln (1-x)}=\lim _{x \rightarrow 0} \frac{f(x)}{x} \cdot \frac{x}{\ln (1-x)}$

$$\lim _{h \rightarrow 0} \frac{1}{h^{2}} f(1-\cos h)=\lim _{h \rightarrow 0} \frac{|1-\cos h|}{h^{2}}=\lim _{h \rightarrow 0} \frac{1-\cos h}{h^{2}}=\lim _{h \rightarrow 0} \frac{2 \sin ^{2}\left(\frac{1}{2} h\right)}{h^{2}}$$
$\sin \left(\frac{1}{2} h\right) \sim \frac{1}{2} h \lim _{h \rightarrow 0} \frac{\frac{1}{2} h^{2}}{h^{2}}=\frac{1}{2}$, 故排除(A)
$$\lim _{h \rightarrow 0} \frac{1}{h^{2}} f(h-\sin h)=\lim _{h \rightarrow 0} \frac{|h-\sin h|}{h^{2}}=\lim _{h \rightarrow 0}\left|\frac{h-\sin h}{h^{3}}\right| \cdot|h|$$ 根据有界量与无穷小的乘积为无穷小, 所以 $\lim _{h \rightarrow 0}\left|\frac{h-\sinh }{h^{3}}\right| \cdot|h|=0$. 故排除(C).

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