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प्रश्न
If \[f\left( x \right) = \begin{cases}\frac{1 - \cos x}{x \sin x}, & x \neq 0 \\ \frac{1}{2} , & x = 0\end{cases}\]
then at x = 0, f (x) is
विकल्प
continuous and differentiable
differentiable but not continuous
continuous but not differentiable
neither continuous nor differentiable
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उत्तर
(a) continuous and differentiable
we have,
\[f\left( x \right) = \begin{cases}\frac{1 - \cos x}{x \sin x}, & x \neq 0 \\ \frac{1}{2} , & x = 0\end{cases}\]
\[\text { Continuity at x } = 0\]
\[(\text { LHL at x }= 0) = {lim}_{x \to 0 -} f(x)\]
\[ = {lim}_{h \to 0} f(0 - h)\]
\[ = {lim}_{h \to 0} f( - h)\]
\[ = {lim}_{h \to 0} \frac{1 - \cos ( - h)}{( - h) \sin ( - h)} \]
\[ = {lim}_{h \to 0} \frac{1 - \cos h}{h \sin h}\]
\[ = {lim}_{h \to 0} 1 - \cos \ h \ {lim}_{h \to 0} \frac{1}{h \sin h}\]
\[ = 1 - \cos(0) . \frac{1}{0 \sin 0} \]
\[ = 0\]
\[(\text { RHL at x }= 0) = {lim}_{x \to 0^+} f(x)\]
\[ = {lim}_{h \to 0} f(0 + h)\]
\[ = {lim}_{h \to 0} f( h)\]
\[ = {lim}_{h \to 0} \frac{1 - \cos (h)}{(h) \sin (h)} \]
\[ = {lim}_{h \to 0} \frac{1 - \cos h}{h \sin h}\]
\[ = {lim}_{h \to 0} 1 - \cos\ h {lim}_{h \to 0} \frac{1}{h \sin h}\]
\[ = 1 - \cos 0 . \frac{1}{0 \sin 0}\]
\[ = 0\]
Hence, f(x)is continuous at x = 0.
For differentiability at x = 0
\[ = {lim}_{h \to 0} \frac{f(0 - h) - f(0)}{0 - h - 0} \]
\[ = {lim}_{h \to 0} \frac{f( - h) - \frac{1}{2}}{- h}\]
\[ = {lim}_{h \to 0} \frac{\frac{1 - \cos( - h)}{- h \sin( - h)} - \frac{1}{2}}{- h}\]
\[ = \frac{1}{h} {lim}_{h \to 0} $\frac{1 - \cos\ h}{h \sin h} - {lim}_{h \to 0} \frac{1}{2}\]
\[ = \frac{1}{2} - 0 = \frac{1}{2}\]
\[\text { RHD at x } = 0 ) = {lim}_{x \to 0^+} \frac{f(x) - f(0)}{x - 0}\]
\[ = {lim}_{h \to 0} \frac{f(0 + h) - f(0)}{0 - h - 0} \]
\[ = {lim}_{h \to 0} \frac{f( h) - \frac{1}{2}}{- h}\]
\[ = {lim}_{h \to 0} \frac{\frac{1 - \cos (h)}{- h \sin(h)} - \frac{1}{2}}{- h}\]
\[ = - \frac{1}{h} {lim}_{h \to 0} \frac{1 - \cos\ h}{h \sin h} - {lim}_{h \to 0} \frac{1}{2}\]
\[ = \frac{1}{2} - 0 = \frac{1}{2}\]
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