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

The standard approach works: multiply numerator and denominator by the reciprocal of the fastest growing term in the denominator to get
$$\begin{align*}
\frac{n^2+3^{2n}}{\left(n^3+3^n\right)^2}&=\frac{n^2+3^{2n}}{n^6+2n^33^n+3^{2n}}\\
&=\frac{n^2+3^{2n}}{n^6+2n^33^n+3^{2n}}\cdot\frac{1/3^{2n}}{1/3^{2n}}\\
&=\frac{\dfrac{n^2}{3^{2n}}+1}{\dfrac{n^6}{3^{2n}}+\dfrac{2n^3}{3^n}+1}\;.
\end{align*}$$
Now take the limit as $n\to\infty$.
