Infinity Limit problem This limit I tried to solve it but I can't I use l'Hospital but it didn't work. Could anyone help me to solve it? 
The limit is :
$$ \lim \limits_{n\to \infty }\frac{n^{2}+3^{n}}{n^{n}+ 2^{n}} =? $$
 A: For sufficiently large $n$, we have $n^2+3^n < 2 \cdot 3^n < 3 \cdot 3^n =3^{n+1} < 3^{2n} < 9^n$ and $n^n + 2^n > n^n$. So $$\frac{n^2+3^n}{n^n+2^n} < \frac{9^n}{n^n} < \frac{9^n}{18^n} = \frac{1}{2^n}$$ for $n \geq 18$. on the other hand, we are dividing two positives, so it is clearly greater than zero. So $$0 =\lim_{n\to\infty} 0 \leq \lim_{n\to\infty} \frac{n^2+3^n}{n^n+2^n} \leq \lim_{n\to\infty} \frac{1}{2^n}=0$$
Moral: First check whether the numerator is growing much faster than the denominator of vice versa. Then the limit will equal 0 or $\pm \infty$. That can you usually prove using inequalities and the Squeeze theorem. If that doesn't work, then you can think about using l'Hôpital. 
A: Hint:
$$
0 \leq \lim \limits_{n\to \infty }\frac{n^{2}+3^{n}}{n^{n}+ 2^{n}} \leq 
\lim \limits_{n\to \infty }\frac{n^{2}+3^{n}}{n^{n}} = 
\lim \limits_{n\to \infty }\left(n^{2-n} + \left( \frac 3n \right)^n\right)
$$
A: The limit is $0$:
$\frac{n^{2}+3^{n}}{n^{n}+ 2^{n}}<\frac{n^{2}+3^{n}}{n^{n}}<n^{2-n}+\left ( \frac{3}{n} \right )^{n}$ and this last item is less than $n^{2-n}+\left ( \frac{3}{4} \right )^{n}$ as soon as $n\geq 4$. Now observe that both of these terms $\to 0$ as $n\to \infty$
