# Solve the limit problem

#### Elize

How am I supposed to solve this problem??

#### romsek

Math Team
As written, it makes no sense.

If the limit were as $n \to \infty$ then divide top and bottom by $n^2$ to obtain

$$\displaystyle \frac{\text{a bunch of stuff over n^2 that goes to zero} + 2n^2}{\text{a few things over n^2 that go to zero} + 7n^2} = \frac 2 7$$

#### SDK

As written, it makes no sense.

If the limit were as $n \to \infty$ then divide top and bottom by $n^2$ to obtain

$$\displaystyle \frac{\text{a bunch of stuff over n^2 that goes to zero} + 2n^2}{\text{a few things over n^2 that go to zero} + 7n^2} = \frac 2 7$$
This doesn't work since the number of things you are neglecting in the numerator is also growing with $n$. It's not hard to prove that the numerator is proportional to a cubic polynomial in $n$ (it even has a closed form), so the limit would be $\infty$.

romsek

#### romsek

Math Team
This doesn't work since the number of things you are neglecting in the numerator is also growing with $n$. It's not hard to prove that the numerator is proportional to a cubic polynomial in $n$ (it even has a closed form), so the limit would be $\infty$.
oops.

Yes, numerator is $$\displaystyle \frac 1 6(n + 3n^2 + 2n^3)$$

#### skipjack

Forum Staff
No, the numerator is ${\large\frac16}(2n^3 + 9n^2 + 13n + 6)$.

#### romsek

Math Team
No, the numerator is ${\large\frac16}(2n^3 + 9n^2 + 13n + 6)$.
if you stop at (n+1) yes, which I suppose I should have done to avoid confusion, but since the limit is in n, I just summed to n.

It's all the same since you let n go to infinity.