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Let $f(x)=\frac{1}{1-|1-x|} .$ Then, what is $\lim _{x \rightarrow 0} f(x)$ equal to
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1
$\begin{aligned} & \text { LHL }=\lim _{h \rightarrow 0} \frac{1}{1-|1-(1-h)|} \\ &=\lim _{h \rightarrow 0} \frac{1}{1-|h|}=\lim _{h \rightarrow 0} \frac{1}{1-h}=1 \\ & \text { RHL }=\lim _{h \rightarrow 0} \frac{1}{1-|1-(1+h)|} \\ &=\lim _{h \rightarrow 0} \frac{1}{1-|-h|}=\lim _{h \rightarrow 0} \frac{1}{1-h}=1 \\ & \therefore \lim _{x \rightarrow 0} f(x)=1 \end{aligned}$
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