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प्रश्न
If \[f\left( x \right) = \begin{cases}\frac{1}{1 + e^{1/x}} & , x \neq 0 \\ 0 & , x = 0\end{cases}\] then f (x) is
पर्याय
continuous as well as differentiable at x = 0
continuous but not differentiable at x = 0
differentiable but not continuous at x = 0
none of these
उत्तर
(d) none of these
we have,
\[(\text{ LHL at x } = 0 )\]
\[ = {lim}_{x \to 0^-} f(x) = {lim}_{h \to 0} f(0 - h) = {lim}_{h \to 0} f( - h)\]
\[ = {lim}_{h \to 0} \frac{1}{1 + e^{1/ - h}}\]
\[ = {lim}_{h \to 0} \frac{1}{1 + \frac{1}{e^{1/h}}} [ {lim}_{h \to 0} \frac{1}{e^{1/h}} = 0] \]
\[ = \frac{1}{1 + 0}\]
\[ = 1\]
\[(\text { RHL at x } = 0) \]
\[ = {lim}_{x \to 0^+} f(x) = {lim}_{h \to 0} f(0 + h)\]
\[ = {lim}_{h \to 0} \frac{1}{1 + e^{1/h}}\]
\[ = \frac{1}{1 + e^{1/0}} = \frac{1}{1 + e^\infty} = \frac{1}{1 + \infty} \]
So, f(x) is not continuous at x = 0
Differentiability at x = 0
\[(\text { LHD at x } = 0 )\]
\[ = {lim}_{x \to 0^-} \frac{f(x) - f(0)}{x - 0}\]
\[ = {lim}_{h \to 0} \frac{f(0 - h) - f(0)}{0 - h - 0}\]
\[ = {lim}_{h \to 0} \frac{f( - h) - 0}{- h}\]
\[ = {lim}_{h \to 0} \frac{\frac{1}{1 + e^{1/ - h}}}{- h}\]
\[ = {lim}_{h \to 0} \frac{\frac{1}{1 + \frac{1}{e^{1/ h}}}}{- h} \]
\[ = {lim}_{h \to 0} \frac{\frac{1}{1 + 0}}{- h} = {lim}_{h \to 0} \frac{1}{- h} = - \infty \]
\[(\text { RHD at x } = 0) \]
\[ = {lim}_{x \to 0^+} \frac{f(x) - f(0)}{x - 0}\]
\[ = {lim}_{h \to 0} \frac{f(0 + h) - f(0)}{0 + h - 0}\]
\[ = {lim}_{h \to 0} \frac{f(h) - 0}{h}\]
\[ = {lim}_{h \to 0} \frac{\frac{1}{1 + e^{1/ h}}}{h} = \infty \]
\[\text{So, f(x) is also not differentiable at x} = 0 . \]
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