# Is it true that $\lim_{(x,y) \rightarrow (0,0)} |\frac{x+\sin y}{y+\sin x}|=1$?

In the calculus 3 textbook there is a problem that asks to show that $$\lim_{(x,y) \rightarrow (0,0)} \frac{x+\sin y}{y+\sin x}$$ does not exist. I found two paths $$y=x$$ and $$y=-x$$ approaching to zero with two different limits $$1$$ and $$-1$$. I also tried other polynomial paths like $$y=x^m$$ or $$x=y^m$$, and the limit is always equal to $$1$$. Hence even if $$\lim_{(x,y) \rightarrow (0,0)} \frac{x+\sin y}{y+\sin x}$$ does not exist, it is expected that in fact $$\lim_{(x,y) \rightarrow (0,0)} |\frac{x+\sin y}{y+\sin x}|=1$$. However I'm not able to prove this. Any comments or ideas would be really appreciated.

• The problem is that within any $\delta$-disc around the origin, you can always find some point $(x, y)$ such that $x + \sin y = 0$ (and $y + \sin x \not= 0$), so the limit fails. – Brian Tung Oct 3 '19 at 21:44
• @BrianTung, Thanks for this clever observation! – student Oct 3 '19 at 22:13

Summing up what you and Brian Tung already wrote: If you approach $$(0,0)$$ along the path $$x= -\sin y$$ you get $$\left| \frac{x+\sin y}{y+\sin x}\right| = \left| \frac0{y-\sin(\sin y)}\right|=0$$. However if you take the path $$x=y$$ you get $$\left| \frac{x+\sin y}{y+\sin x}\right| = \left| \frac{y+\sin y}{y+\sin y}\right| = 1$$.