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

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

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 $\infty$.

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 $\infty$.