Question

Examine the continuity of the function $f\left(x\right)=2x^2-1$ at $x=3$.

Answer 1

We have $f\left(x\right)=2x^2-1$

At $x=3,f\left(x\right)=f\left(3\right)=2(3{)}^2-1=17$

and $\underset{x\to 3}{lim}f\left(x\right)=\underset{x\to 3}{lim}(2x^2-1)=2(3{)}^2-1=17$

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

Thus, $f$ is continuous at $x=3.$