Limit in multivariable calculus

In summary, the limit \ lim_{x\to\,0} \left(\frac{x^{2}}{2x^{2}}\right)\ exists. It is the limit of the function x^{2}/2x^{2}.
  • #1
Gauss M.D.
153
1

Homework Statement



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

[itex]\sqrt{x^{2}+1} - \sqrt{y^{2}-1}[/itex]

[itex]\frac{\sqrt{x^{2}+1} - \sqrt{y^{2}-1}}{x^{2}+y^{2}}[/itex]

The Attempt at a Solution



Tried variable sub:

[itex]\sqrt{x^{2}+1} = a, \sqrt{y^{2}-1} = b[/itex]

[itex]\frac{a - b}{a^{2}-b^{2}}[/itex]

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

Still nasty

Tried polar coordinates:

[itex]\sqrt{1 + r^{2}cosθ} - \sqrt{1 - r^{2}sinθ}/r^{2}[/itex]

But I can't find a way for this limit to exist (which it is supposed to do according to the answers).
 
Physics news on Phys.org
  • #2
I think I got it. Conjugate rule...
 
  • #3
Gauss M.D. said:

Homework Statement



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

[itex]\sqrt{x^{2}+1} - \sqrt{y^{2}-1}[/itex]

[itex]\frac{\sqrt{x^{2}+1} - \sqrt{y^{2}-1}}{x^{2}+y^{2}}[/itex]

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 .
 
  • #4
Yeah, meant to write 1-y^2. But I figured it out anyway :)
 
  • #5
Although I still have something I want to make clear: Say I wanted to show that a particular limit does NOT exist, for example:

[itex]xy^{2}/(x^{2}+y^{4})[/itex]

Is the following valid?

Let y = x:

[itex]x^{3}/(x^{2}+x^{4}) = x/1+x^{2}[/itex]

Which is 0 as x,y -> 0

Now let y = [itex]\sqrt{x}[/itex]

[itex]x^{2}/(2x^{2})[/itex]

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
 
Last edited by a moderator:
  • #6
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.
Gauss M.D. said:
Although I still have something I want to make clear: Say I wanted to show that a particular limit does NOT exist, for example:

[itex]xy^{2}/(x^{2}+y^{4})[/itex]

Is the following valid?

Let y = x:

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

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

Now let y = [itex]\sqrt{x}[/itex]

[itex]x^{2}/(2x^{2}[/itex]
This should have been [itex]\ x^{2}/(2x^{2})\ .[/itex]

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.

[itex]\displaystyle \lim_{x\to\,0} \left(\frac{x^{2}}{2x^{2}}\right)\ [/itex] 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.)
 
  • #7
Yeah that was actually exactly what I meant but got confused in my latex-noobness :S
 

Related to Limit in multivariable calculus

1. What is a limit in multivariable calculus?

A limit in multivariable calculus represents the value a function approaches as its inputs approach a certain point or set of points. It is used to analyze the behavior of functions in multiple dimensions.

2. How is a limit in multivariable calculus different from a limit in single variable calculus?

In single variable calculus, a limit is concerned with the behavior of a function as its input approaches a single value. In multivariable calculus, the limit is concerned with the behavior of a function as its inputs approach a point or set of points in multiple dimensions.

3. What is the importance of limits in multivariable calculus?

Limits in multivariable calculus are important because they allow us to understand the behavior of functions in multiple dimensions and make predictions about their values. They are also essential for determining continuity and differentiability of functions.

4. How are limits in multivariable calculus calculated?

Limits in multivariable calculus are calculated by approaching the given point or set of points along different paths, such as along coordinate axes or along a specific curve. The limit is then evaluated as the inputs approach the specified point or set of points.

5. Can limits in multivariable calculus have more than one value?

No, limits in multivariable calculus can only have one value. This is because as the inputs approach a certain point or set of points, the function can only approach one specific value. If the limit has multiple values, then the limit does not exist.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
567
  • Calculus and Beyond Homework Help
Replies
21
Views
941
  • Calculus and Beyond Homework Help
Replies
5
Views
803
  • Calculus and Beyond Homework Help
Replies
4
Views
613
  • Calculus and Beyond Homework Help
Replies
20
Views
557
  • Calculus and Beyond Homework Help
Replies
6
Views
831
  • Calculus and Beyond Homework Help
Replies
8
Views
859
  • Calculus and Beyond Homework Help
Replies
14
Views
756
Replies
9
Views
796
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
Back
Top