Solution:

The function provided is $f(x)=5 x-3$

Continuity at $x=0$ $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(5 x-3)$ $=5(0)-3$ $=0-3$ $=-3$

Then again, $f(0)=5(0)-3=0-3=-3$

As $\lim _{x \rightarrow 0} f(x)=f(x)$, as a result, $f(x)$ is continuous at $x=0$.

Continuity at $x=-3$,

$\lim _{x \rightarrow-3} f(x)=\lim _{x \rightarrow-3}(5 x-3)=5(-3)-3=-18$

And $f(-3)=5(-3)-3=-18$

As $\lim _{x \rightarrow-3} f(x)=f(x)$, as a result, is continuous at $x=-3$

Continuity at $x=5$,

$\lim _{x \rightarrow 5} f(x)=\lim _{z \rightarrow 5}(5 x-3)$ $=5(5)-3=22$

And $f(5)=5(5)-3=22$

As a result, $\lim _{x \rightarrow 5} f(x)=f(x)$, so, $f(x)$ is continuous at $x=5$.