Question

Find the value of
$a$ for which the function f defined by

$f\left(x\right)=\{\begin{array}{cc}asin\frac{\pi }{2}(x+1)& ,x\le 0\\ \frac{tanx-sinx}{x^3}& ,x>0\end{array}$ is
continuous at $x=0$

Answer 1

For function to be continuous, the limit should exists at $x=0$.

$\underset{x\to 0^{+}}{lim}\frac{tanx-sinx}{x^3}=\underset{x\to 0^{-}}{lim}asin\frac{\pi }{2}(x+1)$

$\underset{x\to 0^{+}}{lim}\frac{tanx-sinx}{x^3}=asin\pi /2=a$

Now using L'Hopital's rule(differentiating $3$ times):

$\underset{x\to 0^{+}}{lim}\frac{4se{c}^{2}xta{n}^{2}x+2se{c}^{4}x+cosx}{6}=a$

$⟹a=\frac{0+2+1}{6}=\frac{1}{2}$