# Limit in multivariable calculus

## Homework Statement

Examine lim (x,y) -> (0,0) of:

$\sqrt{x^{2}+1} - \sqrt{y^{2}-1}$

$\frac{\sqrt{x^{2}+1} - \sqrt{y^{2}-1}}{x^{2}+y^{2}}$

## The Attempt at a Solution

Tried variable sub:

$\sqrt{x^{2}+1} = a, \sqrt{y^{2}-1} = b$

$\frac{a - b}{a^{2}-b^{2}}$

(a -> 1, b -> 1 as x,y -> 0)

Still nasty

Tried polar coordinates:

$\sqrt{1 + r^{2}cosθ} - \sqrt{1 - r^{2}sinθ}/r^{2}$

But I can't find a way for this limit to exist (which it is supposed to do according to the answers).

I think I got it. Conjugate rule.....

SammyS
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Homework Helper
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## Homework Statement

Examine lim (x,y) -> (0,0) of:

$\sqrt{x^{2}+1} - \sqrt{y^{2}-1}$

$\frac{\sqrt{x^{2}+1} - \sqrt{y^{2}-1}}{x^{2}+y^{2}}$

## The Attempt at a Solution

...
Are you sure there's not some typo here?

##\displaystyle \sqrt{y^2-1\,}\ ## is only defined for |y| ≥ 1 .

Yeah, meant to write 1-y^2. But I figured it out anyway :)

Although I still have something I want to make clear: Say I wanted to show that a particular limit does NOT exist, for example:

$xy^{2}/(x^{2}+y^{4})$

Is the following valid?

Let y = x:

$x^{3}/(x^{2}+x^{4}) = x/1+x^{2}$

Which is 0 as x,y -> 0

Now let y = $\sqrt{x}$

$x^{2}/(2x^{2})$

Since I have found a path to the origin with an undefined limit, does that mean I have proven that the original limit does not exist?

Note by SammyS (Mentor):

The following was included a post which was otherwise an exact duplicate of this post.

I'm placing the extra text here because the other post appears in a thread which has been deleted.

Sorry, I meant, is it sufficient to find two curves with DIFFERENT limits?

I.e. let y = x and you have the limit = 0, and y = sqrt(x) and the limit = 1/2

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SammyS
Staff Emeritus
Homework Helper
Gold Member
My first comment: Be careful with parentheses. You have unbalanced parentheses in one of your expressions. One of your other expression has a more serious problem. It needs an additional set of parentheses to make it true.
Although I still have something I want to make clear: Say I wanted to show that a particular limit does NOT exist, for example:

$xy^{2}/(x^{2}+y^{4})$

Is the following valid?

Let y = x:

$x^{3}/(x^{2}+x^{4}) = x/1+x^{2}$
What this says literally is that $\ \displaystyle x^{3}/(x^{2}+x^{4}) = \frac{x}{1}+x^{2}\ .$

What you meant to say (I hope) is $\ \displaystyle x^{3}/(x^{2}+x^{4}) = x/\left(1+x^{2}\right)\,,\$ which is equivalent to $\ \displaystyle x^{3}/(x^{2}+x^{4}) = \frac{x}{1+x^{2}}\ .$
Which is 0 as x,y -> 0

Now let y = $\sqrt{x}$

$x^{2}/(2x^{2}$
This should have been $\ x^{2}/(2x^{2})\ .$

Since I have found a path to the origin with an undefined limit, does that mean I have proven that the original limit does not exist?
If the limit along that second path was indeed undefined, then yes, that would show that the original limit does not exist.

$\displaystyle \lim_{x\to\,0} \left(\frac{x^{2}}{2x^{2}}\right)\$ does exist. What is it?

(It's different than the limit along the path y = x. That does show that the original limit does not exist.)

Yeah that was actually exactly what I meant but got confused in my latex-noobness :S