Using the limit of $ (1+1/n)^n$ to find the limit of $((n^2+2)/(n^2+1)) ^ {3n^2+1/n}$ Use: $ \lim_{n \to \infty} (1+1/n)^n = e$
To find: $ \lim_{n \to \infty} (\frac{n^2+2}{n^2+1}) ^ {3n^2+1/n}$
Any help is appreciated. Not looking for the complete answer since it's looked down upon here. So strong hints would be appreciated. I am lost.

Hint: ${n^2 +2 \over n^2 +1}=1+{1 \over n^2 +1}$
It is convenient to write the exponent as $3n^2 + 1/n= 3(n^2 + 1) -3 + 1/n$ to get:  
$${n^2 +2 \over n^2 +1}^{3n^2 + 1/n}=\left (\left (1+{1 \over n^2 +1} \right )^{n^2+1} \right )^3\left (1+{1 \over n^2 +1} \right )^{-3}\left (1+{1 \over n^2 +1} \right )^{1/n}$$
Now $n^2 +1 \rightarrow +\infty$ as $n \rightarrow +\infty$.
