Let $f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^{2}} & , x

Question

Let $f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^{2}} & , x<0 \ a & , x=0 \ \frac{\sqrt{2}}{\sqrt{16+\sqrt{x-4}}}, & x>0\end{array}ight.$ Determine the value of $a$, if possible, so that the function is continuous at $x=0$

Professor PC Thomas Classes · Answer

As $f$ is continuous at $x=0$\ $f(0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$\ From the $1^{\text {st }}$ part,\ $f(0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{1-\cos 4 x}{x^{2}}$\ $=\lim _{\substack{h \rightarrow 0 \(h>0)}} \frac{1-\cos 4(0-h)}{(0-h)^{2}}=\lim _{h \rightarrow 0} \frac{1-\cos 4 h}{h^{2}}$\ $=\lim _{h \rightarrow 0} \frac{4 \sin 4 h}{2 h}=8 \Rightarrow a=8$.\