If \(\mathrm{f}(x)=\frac{4}{x^4}\left[1-\cos \frac{x}{2}-\cos \frac{x}{4}+\cos \frac{x}{2} \cdot \cos \frac{x}{4}\right]\) is continuous at \(x=0\), then \(\mathrm{f}(0)\) is

\(\begin{aligned}
f(x) & =\frac{4}{x^4}\left[1-\cos \frac{x}{2}-\cos \frac{x}{4}+\cos \frac{x}{2} \cdot \cos \frac{x}{4}\right] \\
& =\frac{4}{x^4}\left[\left(1-\cos \frac{x}{2}\right)-\cos \frac{x}{4}\left(1-\cos \frac{x}{2}\right)\right] \\
& =\frac{4}{x^4}\left[\left(1-\cos \frac{x}{2}\right)\left(1-\cos \frac{x}{4}\right)\right]
\end{aligned}\)
\(\mathrm{f}(x)\) is continuous at \(x=0...[Given]\).
\(\begin{aligned}
\therefore(0) & =\lim _{x \rightarrow 0} \mathrm{f}(x) \\
& =\lim _{x \rightarrow 0} \frac{4}{x^4}\left(1-\cos \frac{x}{2}\right)\left(1-\cos \frac{x}{4}\right) \\
& =4 \lim _{x \rightarrow 0}\left(\frac{2 \sin ^2 \frac{x}{4}}{x^2}\right) \times\left(\frac{2 \sin ^2 \frac{x}{8}}{x^2}\right) \\
& =16 \lim _{x \rightarrow 0}\left(\frac{\sin ^2 \frac{x}{4}}{\frac{x^2}{16}}\right) \times \frac{1}{16} \times \lim _{x \rightarrow 0}\left(\frac{\sin ^2 \frac{x}{8}}{\frac{x^2}{64}}\right) \times \frac{1}{64} \\
& =16 \times \frac{1}{16} \times \frac{1}{64} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{4}}{\frac{x}{4}}\right)^2 \times \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{8}}{\frac{x}{8}}\right)^2 \\
& =\frac{1}{64} \times 1 \times 1 \\
& =\frac{1}{64}
\end{aligned}\)