Question

Find value of $k$, if the function is continuos.
$f\left(x\right)=\{\begin{array}{c}\frac{1-\cos 4x}{x^2}forx<0\\ kforx=0\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}forx>0\end{array}\}\text{at}x=0$

Answer 1

$f\left(x\right)=\{\begin{array}{c}\frac{1-\cos 4x}{x^2}forx<0\\ kforx=0\\ \frac{\sqrt{x}}{\sqrt{16}+\sqrt{x-4}}forx>0\end{array}$

${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}\frac{1-\cos 4x}{x^2}={lim}_{x\to 0^{-}}\frac{2{\sin }^{2}2x}{4x^2}\times 4=2\times 1\times 4=8{lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}\frac{\sqrt{x}}{\sqrt{16}+\sqrt{x-4}}={lim}_{x\to 0^{+}}\frac{\sqrt{x}[\sqrt{16+\sqrt{x}}+4]}{(\sqrt{16+\sqrt{x}}-4)(\sqrt{16+\sqrt{x}}+4)}={lim}_{x\to 0^{+}}\frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16}={lim}_{x\to 0^{+}}\sqrt{16+\sqrt{x}}+4=\sqrt{16+0}+4=4+4=8\therefore f\left(0\right)=8\Rightarrow k=8$