Question

Find $k$ if $f\left(x\right)=\{\begin{array}{cc}k\sin \frac{\pi }{2},& x\le 0\\ \frac{\tan x-\sin x}{x^3},& x>0\end{array}$ is continuous at $x=0$

Answer 1

${lim}_{x\to 0^{+}}\frac{\tan x-\sin x}{x^3}\left(\frac{0}{0}form\right)\Rightarrow \frac{\sin x}{x}.\frac{(\sec x-1)}{x^2}\Rightarrow {lim}_{x\to 0^{+}}\left(\frac{\sec x-1}{x^2}\right)\times {lim}_{x\to 0}\frac{\sin x}{x}\Rightarrow {lim}_{x\to 0}\left(\frac{\sec x-1}{x^2}\right).1as\frac{\sin x}{x}at0=1$

Using L-Hopital

$\Rightarrow {lim}_{x\to 0}\left(\frac{\sec x\tan x}{2x}\right)\Rightarrow \frac{1}{2}{lim}_{x\to 0}\left(\frac{\sin x}{x}\right).{lim}_{x\to 0}\left(\frac{1}{\cos x}\right)\Rightarrow \frac{1}{2}.1.1=\frac{1}{2}$

for a function to be continuous at 0.

${lim}_{x\to 0^{-}}f\left(x\right)=\frac{1}{2}{lim}_{x\to 0^{-}}k\sin \frac{\pi }{2}=\frac{1}{2}(\sin \frac{\pi }{2}=1)k=\frac{1}{2}$