Question

Find the value of a if function f is defined as -
$\left\{asin\frac{\pi }{2}\right(x+1),x\le 0\}\frac{tanx-sinx}{x^3},x>0$
is continuous at x=0

Answer 1

$a\sin \frac{\pi }{2}(x+1)$ $x\le 0$

$=\frac{\tan x-\sin x}{x^3}$ $x>0$

$f(0+h)=\frac{\tan x-\sin x}{x^3}$

$=\frac{\sin x(1-\cos x)}{\cos x.x^3}$

$=\frac{\sin x.2{\sin }^{2}x/2}{x^3}$

$=\underset{x\to 0}{lt}\frac{\sin x}{x}\cdot \frac{2{\sin }^{2}x/2}{\left(\frac{x^2}{4}\right).4}$

$=\frac{1}{2}$

$\therefore \underset{x\to 0}{lt}a\sin \frac{\pi }{2}(x+1)=\frac{1}{2}$

$a=\frac{1}{2}$.