Example 2. Find $\lim _{x \rightarrow \pi / 3} \frac{1-2 \cos x}{\sqrt{3}-2 \sin x} \cdot\left(\frac{0}{0}\right.$ indeterminate form)
Let $u=\operatorname{tg} \frac{x}{2}, x \rightarrow \frac{\pi}{3}$, then $u \rightarrow \sqrt{\frac{3}{3}}$,
$$
\begin{array}{c}
\sin x=\frac{2 u}{1+u^{2}}, \quad \cos x=\frac{1-u^{2}}{1+u^{2}} . \\
\text { Original expression }=\lim _{u \rightarrow \sqrt{3} / 3 \sqrt{3} u^{2}-4 u+\sqrt{3}} \\
u=u-\sqrt{3} / 3 \lim _{u \rightarrow 0} \frac{3 u+2 \sqrt{3}}{\sqrt{3}-2}=-\sqrt{3} .
\end{array}
$$

For expressions containing $\sqrt{a^{2} \pm x^{2}}, \sqrt{x^{2}-a^{2}}$, substitution can be made: $x=a \sin u, x=a \operatorname{tg} u$,
$$
x=a \text { sec } u. \text { For example, } \lim _{x \rightarrow 1} \frac{x-1}{\sqrt{4-x^{2}}-\sqrt{3}}
$$

can be handled this way.
Problems that can be solved using $\lim _{x \rightarrow 0} \frac{\sin x}{x}$ and $\lim _{x \rightarrow \infty}(1+$ $\frac{1}{x})$ are also given as examples.
(Author's affiliation: Hubei Shashi No. 6 Middle School)