Register to reply 
Multivariable Limits 
Share this thread: 
#1
Sep2106, 03:57 PM

P: 29

I was presented with the two following questions:
[tex]\lim_{\substack{x\rightarrow 0\\y\rightarrow 0}} \sin\frac{xy}{xy}[/tex] and [tex]\lim_{\substack{x\rightarrow 0\\y\rightarrow 0}} \sin(\frac{xyz}{xyz})[/tex] 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). 


#2
Sep2106, 09:31 PM

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. 


#3
Sep2206, 05:04 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,327




#4
Sep2206, 12:45 PM

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. 


#5
Sep2206, 01:42 PM

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?)



Register to reply 
Related Discussions  
Multivariable Limits  Calculus & Beyond Homework  5  
Limits of Multivariable Functions  Calculus & Beyond Homework  4  
Multivariable calculus: Limits (2)  Calculus & Beyond Homework  6  
Multivariable Calculus: Limits  Calculus & Beyond Homework  22  
General multivariable limits  General Math  1 