Show $\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = 1$ Could you help me show that
$$\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = 1$$
 A: Divide numerator and denominator by $3^{2n}$, yielding:
$$\frac{\frac{n^2}{3^{2n}}+1}{\left(\frac{n^3}{3^n}+1\right)^2}$$
Prove the numerator and denominator both approach $1$ as $n\to\infty$.
This amounts to showing $\frac{n^2}{3^{2n}}\to 0$ and $\frac{n^3}{3^n}\to 0$ as $n\to\infty$.
A: 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 $n\to\infty$.
A: $\displaystyle \frac{n^2 + 3^{2n}}{(n^3 + 3^n)^2}  = \frac{n^2}{(n^3 + 3^n)^2} + \frac{3^{2n}}{(n^3 + 3^n)^2}$.
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.
A: $$\dfrac{n^2+3^{2n}}{(n^3+3^n)^2}  = \frac{1 + \frac{n^2}{3^{2n}}}{\left( 1 + \frac{n^3}{3^{2n}}\right)^2}$$
As exponential grows more faster than polynomial, you have $1/1$
A: $$\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} =\lim_{n\to\infty}{n^2+3^{2n}\over n^6+2n^33^n+3^{2n}}=\lim_{n\to\infty}{\frac{n^2}{3^{2n}}+1\over\frac{n^6}{3^{2n}}+2\frac{n^3}{3^n}+1}=1$$
