Prove that the function $f(x)=5x-3$ is continuous at $x=0$,$x=-3$,and $x=5$.

The given function is $f(x)=5x-3$.
At $x=0$,$f(0)=5(0)-3=-3$.
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (5x-3) = 5(0)-3 = -3$.
Since $\lim_{x \to 0} f(x) = f(0)$,the function $f$ is continuous at $x=0$.
At $x=-3$,$f(-3)=5(-3)-3=-15-3=-18$.
$\lim_{x \to -3} f(x) = \lim_{x \to -3} (5x-3) = 5(-3)-3 = -18$.
Since $\lim_{x \to -3} f(x) = f(-3)$,the function $f$ is continuous at $x=-3$.
At $x=5$,$f(5)=5(5)-3=25-3=22$.
$\lim_{x \to 5} f(x) = \lim_{x \to 5} (5x-3) = 5(5)-3 = 22$.
Since $\lim_{x \to 5} f(x) = f(5)$,the function $f$ is continuous at $x=5$.