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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Prove that the function f defined by $\text{f(x)}=\begin{cases}\frac{\text{x}}{|\text{x}|+2\text{x}^2},&\text{if x}\neq0\\\text{k},&\text{ if x}=0\end{cases}$ remains discontinuous at x = 0, regardless the choice of k.
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Answer

We have, $\text{f(x)}=\begin{cases}\frac{\text{x}}{|\text{x}|+2\text{x}^2},&\text{if x}\neq0\\\text{k},&\text{ if x}=0\end{cases}$
At x = 0, $\text{L.H.L}=\lim\limits_{\text{x}\rightarrow0^-}\frac{\text{x}}{|\text{x}|+2\text{x}^2}=\lim\limits_{\text{h}\rightarrow0}\frac{(0-\text{h})}{|0-\text{h}|+2(0-\text{h})^2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}+2\text{h}^2}=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}(1+2\text{h})}=-1$
$\text{R.H.L}=\lim\limits_{\text{x}\rightarrow0^+}\frac{\text{x}}{|\text{x}|+2\text{x}^2}=\lim\limits_{\text{h}\rightarrow0}\frac{0+\text{h}}{|0+\text{h}|+2(0+\text{h})^2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}+2\text{h}^2}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}(1+2\text{h})}=1$
And f(0) = k
Since, L.H.L ≠ R.H.L for any value of k.
Hence, f(x) is discontinuous at x = 0 regardless the choice of k.

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