Question

Find the value of k for which
is continuous at x = 0;

Answer 1

Given: $f\left(x\right)=\left\{\begin{array}{l}\frac{1-\cos 4x}{8x^2},\mathrm{when}x\ne 0\\ k,\mathrm{when}x=0\end{array}\right.$

If f(x) is continuous at x = 0, then
$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right) \\ \Rightarrow \underset{x\to 0}{\lim }\frac{1-\cos 4x}{8x^2}=f\left(0\right) \\ \Rightarrow \underset{x\to 0}{\lim }\frac{2{\sin }^{2}2x}{8x^2}=f\left(0\right) \\ \Rightarrow \frac{2}{2}\underset{x\to 0}{\lim }\frac{{\sin }^{2}2x}{4x^2}=f\left(0\right) \\ \Rightarrow \frac{2}{2}\underset{x\to 0}{\lim }{\left(\frac{\sin 2x}{2x}\right)}^2=f\left(0\right) \\ \Rightarrow 1\times 1=f\left(0\right) \\ \Rightarrow k=1\left(\because f\left(0\right)=k\right) \end{gathered}$$

​