Prove that $\lim_{n \to \infty} \frac{3n^2 +n\cos(n)}{n^2+4n} = 3$

Question

Prove that $$\lim_{n \to \infty} \frac{3n^2 +n\cos(n)}{n^2+4n} = 3$$ using the precise definition of the limit.

Without clogging this post up too much with unnecessary algebra I ended up with:

$$\left|\frac{\cos(n)-12}{n+4}\right|< \epsilon$$

This is where I am stuck. I know I am suppose to isolate for $n$, but I don't know how to finish this proof in this case.

Answer 1

$$\frac{3n^2+n\cos n}{n^2+4n}=\frac{\cos n-12}{n+4}+3$$

and $$\left|\frac{\cos n-12}{n+4}\right|<\left|\frac{13}n\right|$$.

You can take $$n>\frac{13}\epsilon$$ to obtain

$$\left|\frac{\cos n-12}{n+4}\right|<\epsilon.$$

Answer 2

$$\left|\frac {\cos n - 12}{n+4}\right|\le \frac {|\cos n|+12}{n+4} \le \frac {12+1}{n+4}$$ as $$|\cos n| \le 1$$ , $ n \in \mathbb{N}$