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What is the value of $\lim _{x \rightarrow 1} \frac{(x-1)^{2}}{|x-1|} ?$
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Let $f(x)=\frac{(x-1)^{2}}{|x-1|}=\left\{\begin{array}{l}(x-1), x \geq 1 \\ -(x-1), x < 1\end{array}\right.$
Now, LHL $\begin{aligned} &=\lim _{h \rightarrow 0} f(1-h) \\=& \lim _{h \rightarrow 0}[-(1-h-1)]=\lim _{h \rightarrow 0} h=0 \end{aligned}$
and $\mathrm{RHL}=\lim _{h \rightarrow 0} f(1+h)$
$\quad=\lim _{h \rightarrow 0}(1+h-1)=\lim _{h \rightarrow 0} h=0$
$\therefore \mathrm{LHL}=\mathrm{RHL}$
$\therefore \quad \lim _{x \rightarrow 0} f(x)=\mathrm{LHL}=\mathrm{RHL}$
$\Rightarrow \quad \lim _{x \rightarrow 0} \frac{(x-1)^{2}}{|x-1|}=0$
Now, LHL $\begin{aligned} &=\lim _{h \rightarrow 0} f(1-h) \\=& \lim _{h \rightarrow 0}[-(1-h-1)]=\lim _{h \rightarrow 0} h=0 \end{aligned}$
and $\mathrm{RHL}=\lim _{h \rightarrow 0} f(1+h)$
$\quad=\lim _{h \rightarrow 0}(1+h-1)=\lim _{h \rightarrow 0} h=0$
$\therefore \mathrm{LHL}=\mathrm{RHL}$
$\therefore \quad \lim _{x \rightarrow 0} f(x)=\mathrm{LHL}=\mathrm{RHL}$
$\Rightarrow \quad \lim _{x \rightarrow 0} \frac{(x-1)^{2}}{|x-1|}=0$
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