Find the limit without L'Hôpital's theorem I'm trying to find $$ \lim_{ n\to\infty} { {n^2+1}\over {n^3+1}} \cdot {\frac {n} {1}}$$
I know the answer is $1$, but I can't remember how my professor found it so simply without using L'Hôpital's theorem.
Could you please show me the shortcut?

We consider $\lim_{ n\to\infty} { {n^2+1}\over n^3+1} \cdot {\frac n 1}$.
To do this, we note that $\lim_{ n\to\infty} { {n^2+1}\over n^3+1} \cdot {\frac n 1} = \lim \frac{n^3 + n}{n^3 + 1} = \lim \frac{n^3\left(1 + \frac{1}{n^2}\right)}{n^3\left( 1 + \frac{1}{n^3}\right)} = \lim \frac{1 + \frac{1}{n^2}}{1 + \frac{1}{n^3}} = 1$
Alternately, we could directly carry out long polynomial division. Divide $n^3 + n$ by $n^3 + 1$. We get $1 + \frac{n-1}{n^3 + 1}$, which clearly has limit $1$.
