Give that $f(x) \begin{cases}=\frac{1-\cos 4 x}{x^2} & ext { if } x 0\end{cases}$ at $x=0$, then $a=$

Question

Give that $f(x) \begin{cases}=\frac{1-\cos 4 x}{x^2} & 	ext { if } x 0\end{cases}$ at $x=0$, then $a=$

Aicharya · Accepted Answer

for continuity at $x=0$ $\begin{aligned} & \lim _{x ightarrow 0} \frac{1-\cos 4 x}{x^2}=a=\lim _{x ightarrow 0} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4} \ & \Rightarrow \lim _{x ightarrow 0} \frac{2 \sin ^2 2 x}{(2 x)^2} 	imes 4=a=\lim _{x ightarrow 0} \frac{\sqrt{x}}{16+\sqrt{x}-16} 	imes(\sqrt{16+\sqrt{x}}+4) \ & \Rightarrow a=8\end{aligned}$

Give that $f(x) \begin{cases}=\frac{1-\cos 4 x}{x^2} & \text { if } x 0\end{cases}$ at $x=0$, then $a=$

Solution