Question
Evaluate $\lim\limits_{\text{x}\rightarrow0}\text{f(x)},$ where $\text{f(x)}=\begin{cases}\frac{|\text{x}|}{\text{x}}, & \text{x} \ne0\\0, &\text{x} = 0\end{cases}.$
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Answer
$\text{L.H.L}=\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^-}\frac{|\text{x}|}{\text{x}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{|0-\text{h}|}{0-\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{+\text{h}}{-\text{h}}=-1\ \cdots(\text{i})$
And,
$\text{R.H.L}=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^+}\frac{|\text{x}|}{\text{x}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0^+}\frac{|\text{x}|}{\text{x}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{|0+\text{h}|}{0+\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}}=1\ \cdots(\text{ii})$
So, $\text{L.H.L}\ne\text{R.H.L}$
$\therefore\ \lim\limits_{\text{x}\rightarrow0}\text{f(x)}$ does not exist.
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^-}\frac{|\text{x}|}{\text{x}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{|0-\text{h}|}{0-\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{+\text{h}}{-\text{h}}=-1\ \cdots(\text{i})$
And,
$\text{R.H.L}=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^+}\frac{|\text{x}|}{\text{x}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0^+}\frac{|\text{x}|}{\text{x}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{|0+\text{h}|}{0+\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}}=1\ \cdots(\text{ii})$
So, $\text{L.H.L}\ne\text{R.H.L}$
$\therefore\ \lim\limits_{\text{x}\rightarrow0}\text{f(x)}$ does not exist.
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