**Solution:**

**Hint: **To check for continuity at a point, check if the function exists at that point and if the limit of the function as x approaches that point is equal to the value of the function at that point.

**Step-by-step Solution:**

**For $x = 3$:**

1. The value of the function at $x = 3$ is:

$f(3) = 2(3^2) - 1 = 2(9) - 1 = 17$

So, $f(3)$ is defined.

2. If $lim_{x\to 3} f(x)$ exists:

$lim_{x\to 3} f(x) = lim_{x\to 3} (2x^2 - 1)$

$\Rightarrow 2(3)^2 - 1 = 18 - 1 = 17$

Therefore, $lim_{x\to 3} f(x)$ exists and is equal to 17.

3. If $lim_{x\to 3} f(x) = f(3)$:

$lim_{x\to 3} f(x) = 17$ and $f(3) = 17$.

Therefore, $lim_{x\to 3} f(x) = f(3)$.

Since all three conditions are met, we can conclude that the function $f(x) = 2x^2 - 1$ is continuous at $x=3$.

**Note:** A polynomial function, such as $f(x) = 2x^2 - 1$, is always continuous everywhere in its domain. The specific point mentioned was just to illustrate the method of proving continuity using the definition and limits.