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$\lim _{x \rightarrow 0} \frac{x\left(e^x-1\right)}{1-\cos x}=$
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$2$
$\begin{aligned} \lim _{x \rightarrow 0} \frac{x\left(e^x-1\right)}{1-\cos x} & =\lim _{x \rightarrow 0} \frac{2 x\left(e^x-1\right)}{4 \cdot \sin ^2 \frac{x}{2}} \\ & =2 \lim _{x \rightarrow 0}\left[\frac{(x / 2)^2}{\sin ^2 \frac{x}{2}}\right]\left(\frac{e^x-1}{x}\right)=2 .\end{aligned}$
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