# 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
• 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