Examine the continuity of the function $f(x)=2 x^{2}-1$ at $x=3$.

Solution:

The provided function is $f(x)=2 x^{2}-1$

Now we will check continuity at $x=3$

$\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3}\left(2 x^{2}-1\right)$

$=2(3)^{2}-1=17$

And $f(3)=2(3)^{2}-1=17$

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