# Find the limit of $\lim_{(x,y)\rightarrow (0,0)} \frac{x-\sin x+y}{x^3 + 6y}$

I have this limit: $$\lim_{(x,y)\rightarrow (0,0)} \frac{x-\sin x+y}{x^3 + 6y}.$$ Taking some different paths I always get the same anwser, $$\frac{1}{6}$$. So I guess that's the limit and I try to prove it using the definition of limit with $$\delta$$ and $$\varepsilon$$ but I can't continue. Any tips/solutions?

• Which paths did you take? Mar 11, 2019 at 10:43
• I tried $x=0$, $y=0$ and $y=x$ Mar 11, 2019 at 11:26

Hint. Recall that as $$x\to 0$$, $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+o(x^5)$$. Now try the path $$y=-x^3/6+ax^5$$ with $$a\not=0$$. What do you get?

• The limit depends on $\alpha$ , so the limit does not exist. But when can I be sure and stop using paths? I tried 3 of them ( $x=0$, $y=0$, $y=x$) and I got $\frac{1}{6}$ every time. So I tried to prove with definition. Can I show somehow with definition that the limit is not $\frac{1}{6}$? Can I find a $\varepsilon$ ? Mar 11, 2019 at 11:27
• Unfortunately checking only a subfamily of paths does not guarantee that the limit exists. Once we have two paths with different limits, then the limit does not exist (this can be proved with $\epsilon-\delta$). Mar 11, 2019 at 11:40
• Yeah I know that checking some paths does not guarantee that the limit exists, but we can assume that this is the limit's solution and then the definition will show us if we are right or not. Am i right? I guess it is not always easy to guess a path that gives us a different limit that the "easy" paths. Mar 11, 2019 at 11:47
• No, it is not always easy to guess a "critical" path, but I don't think that a proof which uses the $\epsilon-\delta$ definition is an easier alternative. Mar 11, 2019 at 12:10