Step 1: Identify the dominant term.
As $n \to \infty$, compare $2 + \sqrt{2} \approx 3.41$ and $2 - \sqrt{2} \approx 0.59$. Since $2 + \sqrt{2} > 1$ and $|2 - \sqrt{2}| < 1$, we have $(2+\sqrt{2})^n \to \infty$ and $(2-\sqrt{2})^n \to 0$.
Step 2: Divide numerator and denominator by $(2+\sqrt{2})^n$.
Let $r = \dfrac{2-\sqrt{2}}{2+\sqrt{2}}$. Then $|r| < 1$, so $r^n \to 0$. The expression becomes: \[ \sqrt{2} \cdot \frac{1 + r^n}{1 - r^n}. \]
Step 3: Take the limit as $n \to \infty$.
Since $r^n \to 0$: \[ \sqrt{2} \cdot \frac{1 + 0}{1 - 0} = \sqrt{2}. \]
Step 4: Verify by computing $r$.
$r = \dfrac{2-\sqrt{2}}{2+\sqrt{2}} = \dfrac{(2-\sqrt{2})^2}{(2+\sqrt{2})(2-\sqrt{2})} = \dfrac{6-4\sqrt{2}}{2} = 3 - 2\sqrt{2} \approx 0.17$. Indeed $|r| < 1$.
Step 5: Confirm $r^n \to 0$.
$r \approx 0.17 < 1$, so $r^n \to 0$ as $n \to \infty$. The computation is valid.
Step 6: State the answer.
The limit is $\sqrt{2}$.
\[ \boxed{\sqrt{2}} \]