Question

Let $f\left(x\right)=\left\{\begin{array}{ll}\frac{1-\cos x}{x^2},\mathrm{when}& x\ne 0\\ 1,\mathrm{when}& x=0\end{array}\right.$
Show that f(x) is discontinuous at x = 0.

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{1-\cos x}{x^2},\mathrm{when}x\ne 0\\ 1,\mathrm{when}x=0\end{array}\right.$

Consider:
$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(\frac{1-\cos x}{x^2}\right) \\ \Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(\frac{2{\sin }^2{\displaystyle \frac{x}{2}}}{x^2}\right) \\ \Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(\frac{2{\sin }^2{\displaystyle \frac{x}{2}}}{4\left({\displaystyle \frac{x^2}{4}}\right)}\right) \\ \Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(\frac{2{\left(\sin {\displaystyle \frac{x}{2}}\right)}^2}{4{\left({\displaystyle \frac{x}{2}}\right)}^2}\right) \\ \Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\frac{2}{4}\underset{x\to 0}{\lim }{\left(\frac{\mathrm{si}{\displaystyle \mathrm{n}}{\displaystyle \frac{x}{2}}}{{\displaystyle \frac{x}{2}}}\right)}^2 \\ \Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\frac{1}{2}·1^2=\frac{1}{2} \end{gathered}$$

Given:
$f\left(0\right)=1$

∴ ​$\underset{x\to 0}{\lim }f\left(x\right)\ne f\left(0\right)$

Thus, f(x) is discontinuous at x = 0.