Example 3. Investigate the continuity of the function

$$
f(x)=\left\{\begin{array}{l}
\frac{x^{2}+7 x+10}{x^{2}-4} \quad \text { for } \quad x<2, x \neq-2 \\
\frac{11}{4} \text { for } \quad x=-2 \\
4 x-3 \quad \text { for } \quad x \geqslant 2
\end{array}\right.
$$
Solution. First, let's investigate the continuity at the point $x=-2$. We have:

$$
\begin{gathered}
\lim _{x \rightarrow-2} f(x)=\lim _{x \rightarrow-2} \frac{x^{2}+7 x+10}{x^{2}-4}=\left(\frac{0}{0}\right)=\lim _{x \rightarrow-2} \frac{(x+2)(x+5)}{(x+2)(x-2)}=-\frac{3}{4} \\
f(-2)=\frac{11}{4} \neq \lim _{x \rightarrow-2} f(x)
\end{gathered}
$$

$x=-2$ is a point of removable discontinuity. If $f(-2)$ were defined as $-\frac{3}{4}$, then $f(x)$ would be continuous at the point $x=-2$.

Now let's consider the point $x=2$:

$$
\begin{aligned}
\lim _{x \rightarrow 2-0} f(x) & =\lim _{x \rightarrow 2-0} \frac{x^{2}+7 x+10}{(x-2)(x+2)}=-\infty \\
\lim _{x \rightarrow 2+0} f(x) & =\lim _{x \rightarrow 2+0}(4 x-3)=5
\end{aligned}
$$

$x=2$ is a point of discontinuity of the second kind. The line $x=2$ is a vertical asymptote (downward, one-sided).