Multivariable Limits: Substitution technique

[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).

 

[0rthodontist]

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.

 

[HallsofIvy]

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.

 

[johndoe3344]

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.

 

[0rthodontist]

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?)