Verify the limit of this succession

I have to verify that

$$\lim_{n\to +\infty} \dfrac{n^2 - n\sin(n)}{3n^2 + \cos(n)} = \dfrac{1}{3}$$

I will write my procedure, asking you to check whether there are mistakes or nonsenses.

$$\left|\dfrac{n^2 - n\sin(n)}{3n^2 + \cos(n)} - \dfrac{1}{3}\right| < \epsilon$$

Now, since $n$ goes to $+$ infinity I could take the absolute value off, but instead I keep it because I need it for the following:

$$\left| \dfrac{-3n\sin(n) - \cos(n)}{9n^2 + 3\cos(n)}\right| < \epsilon$$

$$\dfrac{3n|\sin(n)| + |-\cos(n)|}{9n^2 + 3|\cos(n)|} < \epsilon$$

$$\dfrac{3n+1}{9n^2+3}<\epsilon$$

This now turns out to be $9n^2\epsilon - 3n + 3\epsilon - 1 > 0$$

Which solved gives

$$n = \dfrac{1}{6\epsilon} \pm \dfrac{\sqrt{-108\epsilon^2 + 36\epsilon + 9}}{18\epsilon}$$

Now before getting the intervals,

$$n = \dfrac{1}{6\epsilon} \pm \dfrac{\sqrt{-12\epsilon^2 + 4\epsilon + 1}}{6\epsilon}$$

Can I say now that, for example, $-12\epsilon^2 + 4\epsilon < 5\epsilon$ whence

$$n = \dfrac{1}{6\epsilon} \pm \dfrac{\sqrt{5\epsilon + 1}}{6\epsilon}$$

And eventually, the neighbourhood of infinity is given by

$$n\in\left(\dfrac{1}{6\epsilon}\left(1 + \sqrt{5\epsilon+1}\right) , +\infty\right)$$

Solution 1:

It happens that$$\frac{3n|\sin(n)|+|-\cos(n)|}{9n^2+3|\cos(n)|}\leqslant\frac{3n+1}{9n^2}.$$So, you only need to deal with the inequality $\displaystyle\frac{3n+1}{9n^2}<\varepsilon$. Note that, since $3n+1\leqslant4n$, you can deal only with $\displaystyle\frac4{9n}<\varepsilon$.