Question

Determine the value of the constant $k$ so that the function
$f\left(x\right)=\{\begin{array}{cc}\frac{\sin 2x}{5x},& \text{if}x\ne 0\\ k& ,\text{if}x=0\end{array}$ is continuous at $x=0$

Answer 1

$f\left(x\right)=$ {$\frac{\sin 2x}{5x}$ $x\ne 0$

{$K$ $x\ne 0$

If $f\left(x\right)$ is continuous

$H_{x\to 0}\frac{\sin 2x}{5x}=\frac{0}{0}$ form

Applying $L-$ Hospital's rule

$H_{x\to 0}\frac{2\cos 2x}{5}=\frac{2\cos 0}{5}=2/5$

$H_{x\to 0}f\left(x\right)=f\left(x\right)$ at $x=0$

$\frac{2}{5}=K$

$\therefore$ , $K=2/5$.