Show $\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = 1$ Could you help me show that
$$\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = 1$$

$\displaystyle \frac{n^2 + 3^{2n}}{(n^3 + 3^n)^2}  = \frac{n^2}{(n^3 + 3^n)^2} + \frac{3^{2n}}{(n^3 + 3^n)^2}$.
Since $\frac{n^2}{(n^3 + 3^n)^2} \to 0$ as $n\to \infty$, it's enough to show that $\displaystyle \lim_{n \to \infty} \frac{3^{2n}}{(n^3 + 3^n)^2} = 1$.
To see this, write $ \displaystyle \frac{3^{2n}}{(n^3 + 3^n)^2} = \frac{3^{2n}}{(3^n(\frac{n^2}{3^n} + 1))^2} = \frac{1}{(\frac{n^2}{3^n} + 1)^2}$.
Observing that $\frac{n^2}{3^n} \to 0$ as $n \to \infty$ and the usual statements of continuity, you're done.
