Question

Let $f\left(x\right)=(\begin{array}{c}\frac{1-\cos x}{x^2},\mathrm{if}x\ne 0\\ 1,\mathrm{if}x=0\end{array}$. Which of the following is/are true?

• A. f(x) will be continuous at x = 0 if f(x) = $\frac{1}{2}$ at x = 0
  
• B. f(x) has a removable discontinuity.
• C. $\underset{x\to 0}{\lim }f\left(x\right)\mathrm{exists}.$
  
• D. f(x) has a non-removable discontinuity

Answer 1

Answer: (A), (B), (C)

$\underset{x\to 0}{\lim }f\left(x\right)\mathrm{exists}.$


$f\left(x\right)=(\begin{array}{c}\frac{1-\cos x}{x^2},\mathrm{if}x\ne 0\\ 1,\mathrm{if}x=0\end{array}$$\left(\mathrm{LHL}\mathrm{at}x=0\right)=\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)=\underset{h\to 0}{\lim }\frac{1-\cos \left(-h\right)}{{\left(-h\right)}^2}=\underset{h\to 0}{\lim }\frac{1-\cos \left(h\right)}{h^2}=\underset{h\to 0}{\lim }\frac{2{\sin }^2\left(\frac{h}{2}\right)}{h^2}=\underset{h\to 0}{\lim }\frac{2{\sin }^2\left(\frac{h}{2}\right)}{4{\left(\frac{h}{2}\right)}^2}=\frac{1}{2}\underset{h\to 0}{\lim }{\left(\frac{\sin \left(\frac{h}{2}\right)}{\left(}\right]}^2=\frac{1}{2}\left(\mathrm{RHL}\mathrm{at}x=0\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0+h\right)=\underset{h\to 0}{\lim }\frac{1-\cos h}{h^2}=\underset{h\to 0}{\lim }\frac{2{\sin }^2\left(\frac{h}{2}\right)}{h^2}=\underset{h\to 0}{\lim }\frac{2{\sin }^2\left(\frac{h}{2}\right)}{4{\left(\frac{h}{2}\right)}^2}=\frac{1}{2}\underset{h\to 0}{\lim }{\left(\frac{\sin \left(\frac{h}{2}\right)}{\left(}\right]}^2=\frac{1}{2}\left(\mathrm{LHL}\mathrm{at}x=0\right)=\left(\mathrm{RHL}\mathrm{at}x=0\right)\ne f\left(0\right)\therefore \underset{x\to 0}{\lim }f\left(x\right)\mathrm{exists}.\therefore f\left(x\right)\mathrm{has}\mathrm{removable}\mathrm{discontinuity}\mathrm{at}x=0.f\left(x\right)\mathrm{will}\mathrm{be}\mathrm{continuous}\mathrm{if}f\left(0\right)=\frac{1}{2}$