# Multivariable Limits

by johndoe3344
Tags: limits, multivariable
 P: 29 I was presented with the two following questions: $$\lim_{\substack{x\rightarrow 0\\y\rightarrow 0}} \sin\frac{xy}{xy}$$ and $$\lim_{\substack{x\rightarrow 0\\y\rightarrow 0}} \sin(\frac{xyz}{xyz})$$ I figured I would do a simple substitution: let t = xy for the first one, and the limit becomes as t ->0, sin t /t would approach 1. The answer is right for the first one. Why doesn't the same technique work for the second one? (The answer for the second one is 0).
 Sci Advisor P: 1,253 In the second one are you also meant to have the limit be as z goes to 0? Whenever x, y, and z are all nonzero, then xyz/(xyz) = 1. So sin(xyz/(xyz)) = sin 1. This does not depend on the path that you approach (0, 0, 0) by, so the limit is sin 1.
Math
Emeritus
Thanks
PF Gold
P: 38,708
 Quote by johndoe3344 I was presented with the two following questions: $$\lim_{\substack{x\rightarrow 0\\y\rightarrow 0}} \sin\frac{xy}{xy}$$ and $$\lim_{\substack{x\rightarrow 0\\y\rightarrow 0}} \sin(\frac{xyz}{xyz})$$ I figured I would do a simple substitution: let t = xy for the first one, and the limit becomes as t ->0, sin t /t would approach 1. The answer is right for the first one. Why doesn't the same technique work for the second one? (The answer for the second one is 0).
No, the way you've written it, the lilmit for the second one is not 0. It is, as Orthodontist said, sin 1, exactly like the first.

P: 29

## Multivariable Limits

Sorry. I'm very bad with using latex.

I actually meant [sin (xy )]/(xy)

and likewise for the second one. For the second one, there's also a z->0 which I omitted. Sorry again.
 Sci Advisor P: 1,253 In that case your reasoning is correct and the limit is 1, not 0. Why did you think it is 0? (maybe you're using your calculator to check and you're using degrees instead of radians?)

 Related Discussions Calculus & Beyond Homework 5 Calculus & Beyond Homework 4 Calculus & Beyond Homework 6 Calculus & Beyond Homework 22 General Math 1