Question

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

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{\sin 2x}{5x},\mathrm{if}x\ne 0\\ k,\mathrm{if}x=0\end{array}\right.$
If $f\left(x\right)$ is continuous at x = 0, then
$\underset{\mathrm{x}\to 0}{\lim }f\left(x\right)=f\left(0\right)$
⇒ $\underset{\mathrm{x}\to 0}{\lim }\frac{\sin 2x}{5x}=k$
⇒ $\underset{\mathrm{x}\to 0}{\lim }\frac{2\sin 2x}{5\times 2x}=k$
⇒ $\frac{2}{5}\underset{x\to 0}{\lim }\frac{\sin 2x}{2x}=k$
⇒ $\frac{2}{5}\times 1=k$
⇒ $k=\frac{2}{5}$