Examine the continuity of f(x) at the indicated point $f(x)=\left \{\begin{array}{1 1}\large\frac{1-\cos 2x}{x^2} & \normalsize if\;x\neq 0\\5 & if\;x=0\end{array}\right.$at x=0.

Toolbox:
• A function is said to be discontinuous at $x=a$ if RHL and LHL does not exist.
• It is also discontinuous if RHL $\neq$ LHL.
Step 1:
Given : $f(x)=\left \{\begin{array}{1 1}\large\frac{1-\cos 2x}{x^2} & \normalsize if\;x\neq 0\\5 & if\;x=0\end{array}\right.$at x=0.
The given point is $x=0$
Step 2:
The left hand limit at $x > 0$
Let $\lim\limits_{\large x\to 0^-}f(x)=\lim\limits_{\large x\to 0^-}\large\frac{1-\cos 2x}{x^2}$
But we know $1-\cos 2x=\large\frac{2\sin^2x}{x^2}$
(i.e) $\lim\limits_{\large x\to 0^-}f(x)=\lim\limits_{\large x\to 0^-}\large\frac{2\sin^2x}{x^2}$
$\qquad\qquad\quad\;\;\;=\lim\limits_{\large x\to 0^-}2\big(\large\frac{\sin x}{x}\big)^2$
But we know $\lim\limits_{\large x\to 0^-}\large\frac{\sin\theta}{\theta}$$=1$
On applying limits,
$\lim\limits_{\large x\to 0^-}f(x)=2\times 1$
$\qquad\qquad=2$
Step 3:
The right hand limit at $x < 0$
$\lim\limits_{\large x\to 0^+}f(x)=\lim\limits_{\large x\to 0^+}\large\frac{1-\cos 2x}{x^2}$
$\qquad\quad\quad=\lim\limits_{\large x\to 0^+}\large\frac{2\sin^2x}{x^2}$
$\qquad\quad\quad=\lim\limits_{\large x\to 0^+}2\big(\large\frac{\sin x}{x}\big)^2$
On applying limits,
$\lim\limits_{\large x\to 0^+}f(x)=2\times 1$
$\qquad\qquad=2$
Step 4:
At $x=0$
$f(x)=f(0)=\large\frac{1-\cos 2(0)}{0}$
The function becomes undefined.
So the function is not continuous at $x=0$
It is discontinuous.