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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\frac{2\text{x}+\text{x}^2}{\text{x}}, & \text{x} \neq0\\0,&\text{ x} = 0\end{cases}\text{at x}=0$
✓

Answer

Given,
$\Rightarrow\text{f}\text{(x)}=\frac{2\text{x}+\text{x}^2}{\text{x}},\text{x}>0$
$\Rightarrow\text{f}\text{(x)}=\frac{-2\text{x}+\text{x}^2}{\text{x}},\text{x}>0$
$\Rightarrow\text{f}\text{(x)}=0,\ \text{x}=0$
$\Rightarrow\text{f}\text{(x)}=\text{x+2},\ \text{x}>0$
$\Rightarrow\text{f}\text{(x)}=\text{x-2},\ \text{x}<0$
$\Rightarrow\text{f}\text{(x)}=0,\ \text{x}=0$
We observe
$(\text{LHL at x }=0)$
$\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(0-h)}$
$\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(-h)}=\lim\limits_{\text{h} \rightarrow 0}-\text{h-2}$
$=-2$
$(\text{RHL at x}=0)$
$\lim\limits_{\text{h} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(0+h)}$
$\lim\limits_{\text{h} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(0+h)}$
$\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(h)}=\lim\limits_{\text{h} \rightarrow 0}\text{h+2}$
$=2$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$
Hence, f(x) is discontinuous at x = 0.

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