Question

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

पर्याय

• continuous and differentiable

• differentiable but not continuous

• continuous but not differentiable

• neither continuous nor differentiable

Answer 1

(a) continuous and differentiable 

we have, 

\[f\left( x \right) = \begin{cases}\frac{1 - \cos x}{x \sin x}, & x \neq 0 \\ \frac{1}{2} , & x = 0\end{cases}\]

\[f\left( x \right) = \begin{cases}\frac{1 - \cos x}{x \sin x}, & x \neq 0 \\ \frac{1}{2} , & x = 0\end{cases}\]
\[\text { Continuity at x } = 0\]
\[(\text { LHL at x }= 0) = {lim}_{x \to 0 -} f(x)\]
\[ = {lim}_{h \to 0} f(0 - h)\]
\[ = {lim}_{h \to 0} f( - h)\]
\[ = {lim}_{h \to 0} \frac{1 - \cos ( - h)}{( - h) \sin ( - h)} \]
\[ = {lim}_{h \to 0} \frac{1 - \cos h}{h \sin h}\]
\[ = {lim}_{h \to 0} 1 - \cos \ h \ {lim}_{h \to 0} \frac{1}{h \sin h}\]
\[ = 1 - \cos(0) . \frac{1}{0 \sin 0} \]
\[ = 0\]

\[(\text { RHL at x }= 0) = {lim}_{x \to 0^+} f(x)\]
\[ = {lim}_{h \to 0} f(0 + h)\]
\[ = {lim}_{h \to 0} f( h)\]
\[ = {lim}_{h \to 0} \frac{1 - \cos (h)}{(h) \sin (h)} \]
\[ = {lim}_{h \to 0} \frac{1 - \cos h}{h \sin h}\]
\[ = {lim}_{h \to 0} 1 - \cos\ h {lim}_{h \to 0} \frac{1}{h \sin h}\]
\[ = 1 - \cos 0 . \frac{1}{0 \sin 0}\]
\[ = 0\]

Hence, f(x)is continuous at x = 0.

For differentiability at x = 0

\[(\text { LHD at x }= 0 ) = {lim}_{x \to 0^-} \frac{f(x) - f(0)}{x - 0}\]
\[ = {lim}_{h \to 0} \frac{f(0 - h) - f(0)}{0 - h - 0} \]
\[ = {lim}_{h \to 0} \frac{f( - h) - \frac{1}{2}}{- h}\]
\[ = {lim}_{h \to 0} \frac{\frac{1 - \cos( - h)}{- h \sin( - h)} - \frac{1}{2}}{- h}\]
\[ = \frac{1}{h} {lim}_{h \to 0} $\frac{1 - \cos\ h}{h \sin h} - {lim}_{h \to 0} \frac{1}{2}\]
\[ = \frac{1}{2} - 0 = \frac{1}{2}\]

\[\text { RHD at x } = 0 ) = {lim}_{x \to 0^+} \frac{f(x) - f(0)}{x - 0}\]
\[ = {lim}_{h \to 0} \frac{f(0 + h) - f(0)}{0 - h - 0} \]
\[ = {lim}_{h \to 0} \frac{f( h) - \frac{1}{2}}{- h}\]
\[ = {lim}_{h \to 0} \frac{\frac{1 - \cos (h)}{- h \sin(h)} - \frac{1}{2}}{- h}\]
\[ = - \frac{1}{h} {lim}_{h \to 0} \frac{1 - \cos\ h}{h \sin h} - {lim}_{h \to 0} \frac{1}{2}\]
\[ = \frac{1}{2} - 0 = \frac{1}{2}\]