Verification of $\lim_{n \to \infty} \frac{7n - i}{3n^2 - ni - 2 - i}$

Question

$$\lim_{n \to \infty} \frac{7n - i}{3n^2 - ni - 2 - i}$$

I used L'hopital rule to get it down to $$\lim_{n \to \infty}\frac{7}{5n} = 0$$

Would this be true?

Answer 1

Assuming $i$ is constant (independent of $n$), then yes: the limit is $0$.

I'm not sure how you got $5n$ in the denominator by applying l'Hôpital though, I would expect:

$$\frac{\mbox{d}}{\mbox{d}n}\left(3n^2-ni-2-i\right)=6n-i$$

Answer 2

In this case it can be done a lot simpler than L'Hopital. (Can you apply l'Hopital if the limit is over integers? Is the limit over integers?). Divide both the numerator and the denominator by $n^2$.

Answer 3

Your expression behaves like $$\frac{7}{3n}$$ for large $n$.