# Find the value of 'a' for which the function f defined as $f(x) = \left\{ \begin{array}{l l}asin\frac{\pi}{2}(x+1), & \quad { x \leq 0} \\ \large \frac{\tan\: x-\sin x}{x^3}, & \quad { x > 0} \end{array} \right.$ is continuous at x = 0.

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:
$f(x) = \left\{ \begin{array}{l l}asin\frac{\pi}{2}(x+1), & \quad { x \leq 0} \\ \large \frac{\tan\: x-\sin x}{x^3}, & \quad { x > 0} \end{array} \right.$ is continuous at x = 0.
Since the function is continuous at $x=0$
LHL=RHL
LHL is
$\lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 0^-}a\sin(\large\frac{\pi}{2}$$(x+1) RHL is \lim\limits_{x\to 0^+}f(x)=\lim\limits_{x\to 0^+}\large\frac{\tan x-\sin x}{x^3} Step 2: Since RHL=LHL \lim\limits_{x\to 0^-}a\sin \large\frac{\pi}{2}$$(x+1)=\lim\limits_{x\to 0^+}\large\frac{\tan x-\sin x}{x^3}$
$\lim\limits_{x\to 0^-}f(0)=\lim\limits_{x\to 0^+}f(0)$
$\Rightarrow \lim\limits_{x\to 0}a\sin\big(\large\frac{\pi}{2}+\frac{\pi}{2}$$x)=\lim\limits_{x\to 0}\large\frac{\sin x/\cos x-\sin x}{x^3} \Rightarrow \lim\limits_{x\to 0}a\cos\large\frac{\pi}{2}$$x=\lim\limits_{x\to 0}\large\frac{\sin x(1-\cos x)}{x^3\cos x}$
But $1-\cos x=2\sin^2x/2$
$\Rightarrow \lim\limits_{x\to 0}a\cos\pi/2x=\lim\limits_{x\to 0}\large\frac{\sin x}{x}.\frac{2\sin^2x/2}{x^2/4.4}$
Step 3:
On applying limits we get
$a\cos (0)=1/1.1.1/2=1/2$
(i.e) $a=\large\frac{1}{2}$
Hence the value of $a=\large\frac{1}{2}$
answered Dec 2, 2013