Question

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

Answer 1

The given function is $f\left(x\right)=5x-3$

At $x=0,f\left(0\right)=5\left(0\right)-3=-3$

$\underset{x\to 3}{lim}f\left(x\right)=\underset{x\to 3}{lim}f(5x-3)=5\left(0\right)-3=-3$

$\therefore \underset{x\to 3}{lim}f\left(x\right)=f\left(0\right)$

Therefore, $f$ is continuous at $x=0$

At $x=-3,f(-3)=5(-3)-3=-18$

$\underset{x\to -3}{lim}f\left(x\right)=\underset{x\to -3}{lim}(5x-3)=5(-3)-3=-18$

$\therefore \underset{x\to -3}{lim}f\left(x\right)=f(-3)$

Therefore, $f$ is continuous at $x=-3$

At $x=5,f\left(x\right)=f\left(5\right)=5\left(5\right)-3=25-3=22$

$\underset{x\to 5}{lim}f\left(x\right)=\underset{x\to 5}{lim}(5x-3)=5\left(5\right)-3=22$

$\therefore$ $\underset{x\to 5}{lim}f\left(x\right)=f\left(5\right)$

Therefore, $f$ is continuous at $x=5.$

Hence $f$ is continuous at all the given points (In fact $f$ is continuous for all $R$, Since it is a polynomial )