# For what value of k, is the following function continuous at x = 0 ? $f(x) = \left\{ \begin{array}{l l}\large\frac{1-\cos4x}{8x^2}, & \quad if { x \neq 0 } \\ k, & \quad if { x = 0 } \end{array} \right.$

Toolbox:
• If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
$f(x) = \left\{ \begin{array}{l l}\large\frac{1-\cos4x}{8x^2}, & \quad if { x \neq 0 } \\ k, & \quad if { x = 0 } \end{array} \right.$
Consider the LHL
$\lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 0^-}\large\frac{1-\cos 4x}{8x^2}$
$1-\cos 4x=2\sin^22x$
$\therefore \lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 0^-}\large\frac{2\sin^22x}{8x^2}$
$\qquad\qquad\;\;=\lim\limits_{x\to 0^-}\large\frac{\sin^22x}{4x^2}$
$\qquad\qquad\;\;=\lim\limits_{x\to 0^-}\big(\large\frac{\sin 2x}{2x}\big)^2$
But we know $\lim\limits_{x\to 0}\large\frac{\sin \theta}{\theta}$$=1 Hence \lim\limits_{x\to 0^-}\big(\large\frac{\sin 2x}{2x}\big)^2$$=1$
Step 2:
Consider the RHL
$\lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 0^+}k$
It is given that the function is continuous at $x=0$
$\therefore$ LHL =RHL
$\Rightarrow 1=k$
Hence the value of $k=1$
answered Oct 3, 2013