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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
The function $\text{f(x)=}\begin{cases}\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$
  • A
    is continuous at x = 0
  • ✓
    is not continuous at x = 0
  • C
    is not continuous at x = 0, but can be made continuous at x = 0
  • D
    none of these.
✓

Answer

Correct option: B.
is not continuous at x = 0
Given, $\text{f(x)=}\begin{cases}\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1},\text{x}\neq0\\0,\text{x}=0\end{cases}$

We have

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1}\Bigg)$

if $\text{e}^\frac{1}{\text{x}}=\text{t},$ then

$\text{x}\rightarrow0, \text{t}\rightarrow\infty$

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{t}\rightarrow\infty}\Big(\frac{\text{t}-1}{\text{t}+1}\Big)$

$=\lim\limits_{\text{t}\rightarrow\infty}\Bigg(\frac{1-\frac{1}{\text{t}}}{1+\frac{1}{\text{t}}}\Bigg)=\frac{1-0}{1+0}=1$

Also, f(0) = 0

$\because\ \lim\limits_{\text{x}\rightarrow0}\text{f(x)}\neq\text{f}(0)$

Hence, f(x) is discontinuons at x = 0.

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