Question:
Prove that the function $f(x)=5 x-3$ is continuous at $x=0$, at $x=-3$ and at $x=5$.
Solution:
The given function is $f(x)=5 x-3$
At $x=0, f(0)=5 \times 0-3=-3$
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(5 x-3)=5 \times 0-3=-3$
$\therefore \lim _{x \rightarrow 0} f(x)=f(0)$
Therefore, f is continuous at x = 0
At $x=-3, f(-3)=5 \times(-3)-3=-18$
$\lim _{x \rightarrow-3} f(x)=\lim _{x \rightarrow-3}(5 x-3)=5 \times(-3)-3=-18$
$\therefore \lim _{x \rightarrow-3} f(x)=f(-3)$
herefore, f is continuous at x = −3
At $x=5, f(x)=f(5)=5 \times 5-3=25-3=22$
$\lim _{x \rightarrow 5} f(x)=\lim _{x \rightarrow 5}(5 x-3)=5 \times 5-3=22$
$\therefore \lim _{x \rightarrow 5} f(x)=f(5)$
Therefore, f is continuous at x = 5