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Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity4 Marks
Question
Prove that the function
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}}{|\text{x|+2}\text{x}^2}, &\text{ x}\neq0\\\text{k}, &\text{ x}=0\end{cases}$ 
remains discontinuous at x = 0, regardless the choice of k.
✓

Answer

The given function can be rewritten as,
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}}{\text{x}+2\text{x}^2}, & \text{x} > 0\\\frac{-\text{x}}{\text{x}-2\text{x}^2}, &\text{x} <0\\ \text{k},&\text{x}=0\end{cases}$
$\Rightarrow\text{f}\text{(x)}=\begin{cases}\frac{1}{2\text{x}+1}, &\text{x} > 0\\\frac{1}{2\text{x}-1}, & \text{x} <0\\\text{k},&\text{x} =0\end{cases}$
We observe
$\text{(LHL at x}=0)\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{1}{-2\text{h}-1}=-1$
$\text{(RHL at x}=0)\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{1}{2\text{h}+1}=1$
So,$=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$ Such that
$=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\&\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$
are independent of k.
Thus, f(x) is discontinuous at x = 0,
regardless of the choice of k.

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