Example 6. Find $\lim _{x \rightarrow 0} \frac{\tan x}{x}$.
Solution. Taking into account that $\operatorname{tg} x=\frac{\sin x}{\cos x}$ and $\lim _{x \rightarrow 0} \cos x=\cos 0=1$, based on the properties of limits (1.43) and (1.46) and formula (1.59), we obtain

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

Thus,

$$
\lim _{x \rightarrow 0} \frac{\operatorname{tg} x}{x}=1
$$