# Choosing path to show limit does not exist

I'm trying to show that the limit as $(x,y)$ go to $(0,0)$ for the function $f(x,y) = sin( x + y )/( |x| + |y|)$ does not exist. I initially tried the path $y=2$ and $y=1$, but I don't think I can use these paths because they don't go through the origin. Any thoughts on which path to choose to show that this limit doesn't exist?

• Could I choose $y = x + 1$? This doesn't go through the origin either though. – user180708 Jan 28 '15 at 13:40

## 2 Answers

Try the paths

• $y=x$ with $x>0$; and
• $y=-x$ with $x>0$.

Both paths get arbitrarily close to the origin. Compute the limit restricted to each of those paths (you may need to apply L'Hôpital's rule). If you get two different results, the limit does not exist (because the limit can't depend on the path).

• Yet they will have the same demoninator because of the absolute signs, and the numerator will either be $sin(0)$ or $sin(2x)$ which are both 0. – user180708 Jan 28 '15 at 13:45
• But the limits are different. Do compute them. You'll get a $0/0$ and need to apply L'Hôpital – Luis Mendo Jan 28 '15 at 13:47
• Would I have to use l'hopitals rule twice? Because I have to take the derivative of an absolute value and then I get 0/0 again. – user180708 Jan 28 '15 at 13:55
• Once you have chosen the path you can remove the absolute value, including a minus sign if needed – Luis Mendo Jan 28 '15 at 13:56
• True, but to handle the absolute values easier, it's better to restrict the paths also to x > 0. – Paul Jan 28 '15 at 14:01

Hint: What happens along the curve $y=kx$? Can you think of a curve that approaches the origin and keeps crossing all of these curves over and over while doing so?

• Oh perhaps you mean something like if I take $y= cos(x)$. But that doesn't go through the origin either – user180708 Jan 28 '15 at 14:15
• I was thinking of a spiral into the origin. Easier in polar coordinates. But even a zigzag that bounces between the edges of a wedge as it gets closer to the origin would do. The answer by @LuisMendo is really quite correct. Just show that the limit you obtain by some two paths is different, so that the limit can't exist (must be path independent). – MPW Jan 28 '15 at 15:33