5.22. Let

$$
f(x)=\left\{\begin{array}{l}
e^{3 x}, \text { if } x<0, \\
a+5 x, \text { if } x \geqslant 0
\end{array}\right.
$$

For which choice of the number $a$ will the function $f(x)$ be continuous?
The problem is solved. Let's write down the condition for the continuity of the function at the point \( x=0 \), where it is currently not continuous due to the arbitrariness of the number \( a \). We have

\[
\lim _{x \rightarrow 0-0} f(x)=\lim _{x \rightarrow 0+0} f(x)=f(0)
\]

In this case,

\[
\begin{aligned}
& \lim _{x \rightarrow 0-0} f(x)=\lim _{x \rightarrow 0-0} e^{3 x}=e^{0}=1 ; \\
& \lim _{x \rightarrow 0+0} f(x)=\lim _{x \rightarrow 0-0}(a+5 x)=a ; \\
& f(0)=\left.(a+5 x)\right|_{x=0}=a
\end{aligned}
\]

For the function to be continuous at the point \( x=0 \), the condition \( a=1 \) must be satisfied.