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Continuity and Differentiability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSContinuity and Differentiability4 Marks
Question
$f(x)=\left\{\begin{array}{cl}\frac{x e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, & x \neq 0 \\ 0, & x=0\end{array}, x=0\right.$ Examine the continuity at $x =0$.
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Answer

Given function $f(x)=\left\{\begin{array}{cc}\frac{x e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, & x \neq 0 \\ 0, & x=0\end{array}\right.$, is continuous at $x=0$
$\therefore$ Value of $\text{R.H.L.}$
$\lim _{x \rightarrow 0^{+}} f(x) =\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} \frac{(0+h) e^{\frac{1}{0+h}}}{1+e^{\frac{1}{0+h}}}$
$ =\lim _{h \rightarrow 0} \frac{h e^{\frac{1}{h}}}{e^{\frac{1}{h}}\left[\left(\frac{1}{e^{\frac{1}{h}}}+1\right)\right]}$
$ =\lim _{h \rightarrow 0} \frac{h}{\frac{1}{e^{1 / h}}+1}=\frac{0}{\frac{1}{\infty}+1}=\frac{0}{0+1}$
$ =\frac{0}{1}=0$
value of $\text{L.H.L.}$
$\lim _{x \rightarrow 0^{-}} f(x) =\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} \frac{(0-h) e^{\frac{1}{0-h}}}{1+e^{\frac{1}{0-h}}}$
$ =\lim _{h \rightarrow 0} \frac{-h e^{-1 / h}}{1+e^{-1 / h}}=\lim _{h \rightarrow 0}\left\{\frac{(-h) \frac{1}{e^{1 / h}}}{1+\frac{1}{e^{1 / h}}}\right.$
$\qquad \quad=\frac{0}{1}=0$
 value of $ f(0)=0$
$\because \lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(0-h)=f(0)$
$\therefore $ hence function is continuous at $ x=0$

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