Question

If $f\left(x\right)=(\begin{array}{c}\frac{1-\cos x}{x\sin x},x\ne 0\\ \frac{1}{2},x=0\end{array}$ then at x = 0, f(x) is

• A. continuous
• B. not continuous
• C. differentiable
• D. not differentiable

Answer 1

Answer: (A), (C)

The correct options are
A continuous
C differentiable

We have,
$f\left(x\right)=(\begin{array}{c}\frac{1-\cos x}{x\sin x},x\ne 0\\ \frac{1}{2},x=0\end{array}\Rightarrow f(x)=(\begin{array}{c}\frac{2{\sin }^2\left(x/2\right)}{x\left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)},x\ne 0\\ \frac{1}{2},x=0\end{array}\Rightarrow f\left(x\right)=\left(\begin{array}{c}\frac{\tan \frac{x}{2}}{x},x\ne 0\\ \frac{1}{2},x=0\end{array}\right(\mathrm{LHL}\mathrm{at}x=0)=\underset{x\to 0^{-}}{\lim }f(x)=\underset{h\to 0}{\lim }f(0-h)=\underset{h\to 0}{\lim }\frac{\tan \frac{-h}{2}}{-h}=\frac{1}{2}\underset{h\to 0}{\lim }\frac{\tan \frac{-h}{2}}{-\frac{h}{2}}=\frac{1}{2}(\mathrm{RHL}\mathrm{at}x=0)=\underset{x\to 0^{+}}{\lim }f(x)=\underset{h\to 0}{\lim }f(0+h)=\underset{h\to 0}{\lim }\frac{\tan \frac{h}{2}}{h}=\frac{1}{2}\underset{h\to 0}{\lim }\frac{\tan \frac{h}{2}}{\frac{h}{2}}=\frac{1}{2}(\mathrm{LHL}\mathrm{at}x=0)=(\mathrm{RHL}\mathrm{at}x=0)=f(0)\therefore f(x)\mathrm{is}\mathrm{continuous}\mathrm{at}x=0.(\mathrm{LHD}\mathrm{at}x=0)=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{h\to 0}{\lim }\frac{f\left(0-h\right)-f\left(0\right)}{0-h-0}=\underset{h\to 0}{\lim }\frac{\frac{\tan \left(-\frac{h}{2}\right)}{-h}-\frac{1}{2}}{-h}=\underset{h\to 0}{\lim }\frac{\tan \left(-\frac{h}{2}\right)+\frac{h}{2}}{h^2}=\underset{h\to 0}{\lim }\frac{\left(\frac{-h}{2}+\frac{{\left(\frac{-h}{2}\right)}^3}{3}+\frac{2{\left(\frac{-h}{2}\right)}^5}{15}....\right)+\frac{h}{2}}{h^2}=\underset{h\to 0}{\lim }\frac{\frac{{\left(\frac{-h}{2}\right)}^3}{3}+\frac{2{\left(\frac{-h}{2}\right)}^5}{15}....}{h^2}=\underset{h\to 0}{\lim }\frac{\frac{-h}{8}}{3}+\frac{2\frac{-h^3}{32}}{15}....=0(\mathrm{RHD}\mathrm{at}x=0)=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{h\to 0}{\lim }\frac{f(0+h)-f\left(0\right)}{0+h-0}=\underset{h\to 0}{\lim }\frac{\frac{\tan \left(\frac{h}{2}\right)}{h}-\frac{1}{2}}{h}=\underset{h\to 0}{\lim }\frac{\tan \left(\frac{h}{2}\right)-\frac{h}{2}}{h^2}=\underset{h\to 0}{\lim }\frac{\left(\frac{h}{2}+\frac{{\left(\frac{h}{2}\right)}^3}{3}+\frac{2{\left(\frac{h}{2}\right)}^5}{15}....\right)-\frac{h}{2}}{h^2}=\underset{h\to 0}{\lim }\frac{\frac{{\left(\frac{h}{2}\right)}^3}{3}+\frac{2{\left(\frac{h}{2}\right)}^5}{15}....}{h^2}=\underset{h\to 0}{\lim }\frac{\frac{h}{8}}{3}+\frac{2\frac{h^3}{32}}{15}....=0(\mathrm{LHD}\mathrm{at}x=0)=(\mathrm{RHD}\mathrm{at}x=0)\therefore f(x)\mathrm{is}\mathrm{differentiable}\mathrm{at}x=0$