Multivariable limits (NOT TO THE ORIGIN)

[mirajshah]

Homework Statement

Hi everyone! I'm pretty good with multivariable limits, but this one has me stumped:
Find the limit or show that it does not exist:
<br /> \underset{\left(x,y,z\right)\rightarrow\left(1,-1,1\right)}{\lim}\frac{yz+xz+xy}{1+xyz}<br />

Homework Equations

The Attempt at a Solution

I could not work with polar coordinates here because there is no easy way to find a value that ρ approaches (the point is (1, -1, 1).
I found it difficult to prove that the limit didn't exist, as in this case the point is in 4-d space and one can only approach it from various 3-D spaces, which I simply could not visualize how to do.
Help please!

 

[sunjin09]

simply take the limit along one axis, say x axis, the limit already blows up

 

[mirajshah]

Read the problem buddy. The limit is to be taken approaching the point (1, -1, 1), which is a point that doesn't lie on any of the axes.

 

[Dick]

The numerator approaches 0 and the denominator doesn't approach 0. Isn't that enough to tell you about the behavior of the limit?

 

[mirajshah]

Doesn't the denominator approach: \left(1+\left(1\times-1\times1\right)\right)=\left(1+\left(-1\right)\right)=\left(1-1\right)=0?

sunjin09, I realize I was a little rude with my response and I'm sorry, it's been a tough week. Thanks for the responses guys! I appreciate it.

Quick note: plugging into Wolfram Alpha yields nothing, so I don't have an answer we can cross-reference against. Sorry!

 

[HallsofIvy]

Yes, the denominator approaches 0. And the numerator doesn't. That was Dick's point.

 

[Dick]

Yes, the denominator approaches 0. And the numerator doesn't. That was Dick's point.

Yes, I had them backwards, sorry.

 

[mirajshah]

Oh my god, the limit is -∞? I'm so sorry guys, I feel like a real idiot. Thanks for the help!

 

[Dick]

Oh my god, the limit is -∞? I'm so sorry guys, I feel like a real idiot. Thanks for the help!

You are welcome! But I wouldn't describe it that way. The denominator doesn't have a definite sign. It could be either +∞ or -∞ depending on how you approach it.

 

[sunjin09]

Doesn't the denominator approach: \left(1+\left(1\times-1\times1\right)\right)=\left(1+\left(-1\right)\right)=\left(1-1\right)=0?

sunjin09, I realize I was a little rude with my response and I'm sorry, it's been a tough week. Thanks for the responses guys! I appreciate it.

Quick note: plugging into Wolfram Alpha yields nothing, so I don't have an answer we can cross-reference against. Sorry!

No worries. I should've been more accurate in my wording.