# Help with this Multivariable Limit

• guyvsdcsniper
Why over-complicate things?In summary, the problem is that -x is factored into sqrt(x)*sqrt(y) but the sign is different in the numerator and denominator.f

#### guyvsdcsniper

Homework Statement
Find the limit
Relevant Equations
lim(x,y)->(0,0)

I do not understand how they got the -x in the numerator to turn into a sqrt(x) when factoring to solve this multivariable function. Could some help me understand?

Last edited:
That is not the same problem. There is a sign difference in the first term of the numerator.

• guyvsdcsniper
That is not the same problem. There is a sign difference in the first term of the numerator.
I missed that. I need to edit my post. The 1st problem in the 2nd image is what I am questioning. How did they get -x to become sqrt(x)?

I missed that. I need to edit my post. The 1st problem in the 2nd image is what I am questioning. How did they get -x to become sqrt(x)?
In the work shown, ##\sqrt{xy} - x## was factored into ##\sqrt x(\sqrt y - \sqrt x)##.

Clear?

• Delta2 and guyvsdcsniper
In the work shown, ##\sqrt{xy} - x## was factored into ##\sqrt x(\sqrt y - \sqrt x)##.
Yea that is what's getting me. I get separating sqrt(xy) into sqrt(x)*sqrt(y). I just don't see how the -x turns into sqrt(x) when factoring.

In the work shown, ##\sqrt{xy} - x## was factored into ##\sqrt x(\sqrt y - \sqrt x)##.

Clear?
Oh wait. The -x is technically -x^1 and sqrt of x is technically x^1/2. so if pull a x^1/2 you are left with a x^1/2. Is that correct?

• Delta2
Oh wait. The -x is technically -x^1 and sqrt of x is technically x^1/2. so if pull a x^1/2 you are left with a x^1/2. Is that correct?
Why over-complicate things?
I get separating sqrt(xy) into sqrt(x)*sqrt(y).
But you don't get separating sqrt(xx) into sqrt(x)*sqrt(x)? Edit: or ## \dfrac {x}{\sqrt x} = \sqrt x ##?

Note that we must be sure that x (and y) are non-negative for these manipulations.

• Delta2
Oh wait. The -x is technically -x^1 and sqrt of x is technically x^1/2. so if pull a x^1/2 you are left with a x^1/2. Is that correct?
Yes. You could also try multiplying out the factorization @Mark44 showed you and see that you recover what you started with.