I've written a proof for the following limit -it may seem a little bit long but that's just because I've included every small step I made to make it clear-:
$$\lim_{(x,y)\to(0,3)} \frac{y^2 + cos (\pi - x)}{y} = \frac{8}{3}$$
I'm unsure of it's correctness, and there are certain points where I have a few doubts.
For an $\epsilon > 0$ I need to find a $\delta>0$ such that:
$|\frac{y^2 + cos (\pi - x)}{y} - \frac{8}{3}| < \epsilon$
When the distance from $(x,y)$ to the point $(0,3)$ is less than $\delta$. From this I know that $|x|$ and $|y-3|$ are both less than $\delta$. So I start from the inequality above:
$$|\frac{y^2 + cos (\pi - x)}{y} - \frac{8}{3}| =$$
$$|\frac{y^2}{y} + \frac{cos (\pi - x)}{y} - \frac{8}{3}| =$$
$$|y - 3 + \frac{cos (\pi - x)}{y} - \frac{8}{3} + 3| =$$
$$|y - 3 + \frac{cos (\pi - x)}{y} + \frac{1}{3}| =$$
$$|y - 3 + \frac{3 cos (\pi - x) + y}{3y}| =$$
$$|y - 3 + \frac{3 cos (\pi - x) + y + 3 - 3}{3y}| =$$
$$|y - 3 + \frac{y - 3}{3y} + \frac{3 cos (\pi - x) + 3}{3y}| \leq$$
$$|y - 3| + \frac{|y - 3|}{3|y|} + \frac{3}{3|y|} \cdot |cos (\pi - x) + 1|$$
Here I was suggested that if I consider $\delta < 1$ then $||(x,y)-(0,3)||<1$ and as $|y-3|<1$ then $|y| \geq 2$ and $\frac{1}{|y|} \leq 2$, so $\frac{1}{3|y|}\leq \frac{2}{3}$. This is something that I don't understand how and why, could you explain this to me?
Anyway I kept going this way:
$$|y - 3| + \frac{2|y - 3|}{3} + 2|cos (\pi - x) + 1|$$
And now I have to consider $|cos (\pi - x) + 1|$. So I need to find an alternative $\epsilon'$ and $\delta'$ so I can bound that expression to $0$. I can choose that alternative $\epsilon'$ to be half of the first one, and it will follow that $|x|$ is less than the alternative $\delta'$, in the end I need the original $\delta$ to be less than this alternative one. I need help with this auxiliar step.
$$|y - 3| + \frac{2|y - 3|}{3} + 2|cos (\pi - x) + 1| < \delta + \frac{2 \delta}{3} + \frac{\epsilon}{2}<\epsilon$$
$$\delta < \frac{3 \epsilon}{10}$$
And I should choose $\delta$ to be the minimum of $\{1, \delta', \frac{3 \epsilon}{10}\}$
That's it. To recap I would need help in checking that the proof is alright and then an explanation on the part dealing with choosing $\delta < 1$ and the part of $|cos (\pi - x) + 1|$.
Thank you so much!