Example 5. Find $\lim _{x \rightarrow 2}(2-x) \tan \frac{\pi}{4} x$.
Solution. When $x \rightarrow 2$, we have an indeterminate form of $0 \cdot \infty$. Setting $x=2-\alpha$ and taking the limit, we find

$$
\begin{aligned}
& \lim _{x \rightarrow 2}(2-x) \operatorname{tg} \frac{\pi}{4} x=\lim _{\alpha \rightarrow 0} \alpha \operatorname{tg} \frac{\pi}{4}(2-\alpha)=\lim _{\alpha \rightarrow 0} \alpha \operatorname{tg}\left(\frac{\pi}{2}-\frac{\pi}{4} \alpha\right)= \\
& =\lim _{\alpha \rightarrow 0} \alpha \operatorname{ctg} \frac{\pi}{4} \alpha=\lim _{\alpha \rightarrow 0} \alpha \frac{\cos \frac{\pi}{4} \alpha}{\sin \frac{\pi}{4} \alpha}=\lim _{\alpha \rightarrow 0} \frac{\alpha}{\sin \frac{\pi}{4} \alpha} \cos \frac{\pi}{4} \alpha= \\
& =\lim _{\alpha \rightarrow 0} \frac{\frac{\pi}{4} \alpha}{\frac{\pi}{4} \sin \frac{\pi}{4} \alpha} \cdot \lim _{\alpha \rightarrow 0} \cos \frac{\pi}{4} \alpha=\frac{1}{\frac{\pi}{4}} \cdot 1=\frac{4}{\pi}
\end{aligned}
$$