The function ${\text{⨍ (x)=(sin2x)}}^{{\tan }^2\text{2x}}$ is not defined at $\mathrm{x}=\mathrm{\pi }/4$. The value of $\text{⨍ (π/4)}$ so that $\text{⨍ }$ is continuous at $\mathrm{x}=\mathrm{\pi }/4$ is

$$\begin{gathered} \mathrm{f}\text{ is continuous at }x=\frac{\mathrm{\pi }}{4}\text{, if }\underset{x\to \pi /4}{lim} f\left(x\right)=f\left(\frac{\mathrm{\pi }}{4}\right) \\ \mathrm{Let} \mathrm{L}==\underset{x\to \pi /4}{lim} {\left(\sin 2x\right)}^{{\tan }^{2}2x} \\ \begin{array}{l}\Rightarrow \log L=\underset{x\to \pi /4}{lim} {\tan }^{2}2x\left(\log \sin 2x\right)\\ =\underset{x\to \pi /4}{lim} \frac{\log \sin 2x}{{\cot }^{2}2x}\left(\frac{\mathrm{\infty }}{\mathrm{\infty }}\right)\\ =\underset{x\to \pi /4}{lim} \frac{2\cot 2x}{-2\cot 2x\cos e{c}^{2}2x\cdot 2}=-\frac{1}{2}\\ \text{ or }L=e^{-1/2}\\ \therefore f\left(\frac{\mathrm{\pi }}{4}\right)=e^{-1/2}=1/\sqrt{e}\end{array} \end{gathered}$$