Question

# Evaluate $\underset{x\to 0}{\mathrm{lim}}f\left(x\right)$, where $f\left(x\right)=\left\{\begin{array}{ll}\frac{\left|x\right|}{x},& x\ne 0\\ 0,& x=0\end{array}\right\$

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Solution

## $f\left(x\right)=\left\{\begin{array}{ll}\frac{\left|x\right|}{x},& x\ne 0\\ 0,& x=0\end{array}\right\\phantom{\rule{0ex}{0ex}}\mathrm{LHL}:\phantom{\rule{0ex}{0ex}}\underset{x\to {0}^{-}}{\mathrm{lim}}\left(\frac{\left|x\right|}{x}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Let}x=0-h,\mathrm{where}h\to 0.\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}\left(\frac{\left|0-h\right|}{-h}\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{h}{-h}\right)=-1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{RHL}:\phantom{\rule{0ex}{0ex}}\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}=\underset{x\to {0}^{+}}{\mathrm{lim}}\left(\frac{\left|x\right|}{x}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Let}x=0+h,\mathrm{where}h\to 0.\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{\mathrm{lim}}\left(\frac{\left|0+h\right|}{0+h}\right)=1\phantom{\rule{0ex}{0ex}}\mathrm{LHL}\ne \mathrm{RHL}\phantom{\rule{0ex}{0ex}}\underset{x\to 0}{\mathrm{Thus},\mathrm{lim}}f\left(x\right)\mathrm{does}\mathrm{not}\mathrm{exist}.$

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