Question

Find the value of ${}^{\prime }{a}^{\prime }$ for which
$f\left(x\right)=\{\begin{array}{c}k\sin \frac{\pi }{2}(x+1),x\le 0\\ \frac{\tan x-\sin x}{x^3},x>0\end{array}$ is continuous at $x=0$

Answer 1

$f\left(x\right)=\{\begin{array}{c}a\sin \pi /2(x+1)\\ \frac{\tan x-\sin x}{x^3}\end{array}$

${lim}_{x\to 0}a\sin \pi /2(x+1)=a\sin \pi /2$

${lim}_{x\to 0}\frac{\tan x-\sin x}{x^3}=\frac{\sin x(\frac{1}{\cos x}-1)}{x^3}\Rightarrow \frac{\sin x}{x}.\frac{2{\sin }^{2}x/4}{4x^2/4}$

${lim}_{x\to 0}\frac{\tan x-\sin x}{x^3}=\frac{2}{4}=\frac{1}{2}$

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