# 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?

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$\frac{n^3 + n }{n^3 + 1} = \frac{1 + \frac{1}{n^2}}{1+ \frac{1}{n^3}}.$ –  Tim Duff Jul 30 '12 at 5:50
@timduff That doesnt look like the same forumla I put? –  Deekor Jul 30 '12 at 5:54
What happens to $\frac{1}{n}$ as $n$ tends to infinity? –  Tim Duff Jul 30 '12 at 5:57
@timduff it becomes 0, but my formula says n/1 not 1/n –  Deekor Jul 30 '12 at 6:00
@markdominus you got it right –  Deekor Jul 30 '12 at 6:01

First let's expand $$\lim_{ n\to\infty} { {n^2+1}\over n^3+1} \cdot {\frac n 1\phantom{n^2}}$$ using regular algebra:

$$\lim_{ n\to\infty} { {n^3+n}\over n^3+1}$$

There's a standard trick for dealing with rational functions (that is, with quotients of polynomials). One way of looking at it is that you can divide the top and bottom by $n^k$ where $k$ is the largest power that appears in the numerator or denominator; here that is $n^3$:

$$\lim_{ n\to\infty} { {n^3+n}\over n^3+1} \cdot {\frac1{n^3}\over\frac1{n^3}\\ \lim_{ n\to\infty} { {1+\frac1{n^2}}\over 1+\frac1{n^3}} }$$

As $n$ gets very large, $\frac1{n^2}$ and $\frac1{n^3}$ become insignificant compared with 1, so we are left with $\lim_{ n\to\infty} \frac11 = 1$. This is what Tim Duff did in the comments above.

But the short version of the same thing is to observe that it lets us disgregard all but the largest terms of the polynomials in the numerator and denominator. If the numerator has the largest power of $n$, the limit is $\infty$; if the denominator has the largest power of $n$, the limit is 0, and if the powers are the same, as they are here, then you have $an^k\over bn^k$ and the limit is $\frac ab$.

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${ {n^2+1}\over {n^3+1}} \cdot {\frac {n} {1}} = { {n^3+n}\over {n^3+1}} = { {n^3+1-1+n}\over {n^3+1}} = { {n^3+1}\over {n^3+1}} + { {-1+n}\over {n^3+1}} = 1 + { {-1+n}\over {n^3+1}}$.

The limit of the right hand term is zero, so the overall limit is $1$.

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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 $1$.

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When dealing with rational functions in $n$, if $n\to \infty$, then the terms which matter are of highest degree in both numerator and denominator, so neglect other terms and then your limit becomes very simple.
Here in your problem $$\lim_{n\to\infty}\frac{n^3+n}{n^3+1},$$ neglect $n$ in numerator as we have $n^3$ with it and neglect $1$ in denominator as we have $n^3$ with it, hance the limit becomes $$\lim_{n\to\infty}\frac{n^3}{n^3}=1$$You can really do it in your mind.
$\displaystyle \lim_{n \to \infty }\frac{n^2+1}{n^3+1}\cdot \frac{n}{1} =\lim_{n \to \infty }\frac{n^3+n}{n^3+1} =\lim_{n \to \infty }\frac{1+1/n^2}{1+1/n^3} =\frac{1+1/\infty}{1+1/\infty} =\frac{1+0}{1+0} =1$