Find limit of multi variable function

In summary, the conversation discusses the calculation of a limit using polar coordinates and the conclusion that the limit does not exist. One person suggests checking if the limit depends on θ, while another suggests considering multiple paths of approach. However, the second person's rule of thumb is deemed inaccurate by another person due to the differing limits for certain values.
  • #1
ananonanunes
18
6
Homework Statement
Given the continuous function ##g\colon \mathbb{R} \rightarrow \mathbb{R}## so that ##g(1)=7##, consider the function defined in ##\mathbb{R}^{2}_{\backslash(1,0)}## by$$f(x,y)=\frac {g(x)(x-1)^2y}{2(x-1)^4+y^2}.$$

Say whether the limit ##\lim_ {(x,y) \rightarrow (1,0)} {f(x,y)}## exists.
Relevant Equations
_
This is what I did: $$\lim_ {(x,y) \rightarrow (1,0)} {\frac {g(x)(x-1)^2y}{2(x-1)^4+y^2}}=\lim_ {(x,y) \rightarrow (1,0)} {g(x)y\frac {(x-1)^2}{2(x-1)^4+y^2}}$$ I know that ##\lim_ {(x,y) \rightarrow (1,0)} {g(x)y}=0## and that ##\frac {(x-1)^2}{2(x-1)^4+y^2}## is limited because ##0\leq (x-1)^2\leq 2(x-1)^4\leq 2(x-1)^4+y^2\Rightarrow 0 \leq\frac{(x-1)^2}{2(x-1)^4+y^2}\leq1##, so I concluded that the limit exists and it is 0.
I know this is wrong and I can understand that the limit cannot exist because if you calculate the limit for ##y=(x-1)^2## you get a solution different than 0. But I thought that having a function whose limit is 0 times a function that is limited was enough to conclude that the limit of their multiplication was 0. Is this wrong?
 
Physics news on Phys.org
  • #2
I did not know that
$$(x-1)^2\leq 2(x-1)^4$$
did I miss something ?
 
  • Love
  • Like
Likes malawi_glenn and ananonanunes
  • #3
BvU said:
I did not know that
$$(x-1)^2\leq 2(x-1)^4$$
did I miss something ?
You're right! I totally forgot about the numbers between 0 and 1. Thanks for the help
 
  • #4
Set [itex]x = 1 + r\cos \theta[/itex], [itex]y = r \sin \theta[/itex] and consider the limit [itex]r \to 0[/itex]. Does it depend on [itex]\theta[/itex]?

EDIT This is not sufficient to conclude that the limit exists, as the example I give below shows.

Instead consider what happens if you approach (1,0) along the path [itex](1 + t, At^2)[/itex] for arbitrary [itex]A[/itex]. If the limit exists, then the result should be independent of [itex]A[/itex].
 
Last edited:
  • #5
pasmith said:
Set [itex]x = 1 + r\cos \theta[/itex], [itex]y = r \sin \theta[/itex] and consider the limit [itex]r \to 0[/itex]. Does it depend on [itex]\theta[/itex]?
It does. So if I have a two variable function and I want to calculate the limit for (x,y) to (0,0) I can just replace x and y with the polar coordinates and check whether it depends on \theta and if it does, there is no limit?
 
  • #6
Maybe I shouldn't, but I'd suggest you consider 2 or more random paths of approach. If you get the same limit on both, the limit likely exists. Good if you're stuck in an exam with little time left. Easy choices: along ##y=x## along ##(0,y) ##or ##(x,0)##
 
  • Like
Likes ananonanunes and BvU
  • #7
ananonanunes said:
It does. So if I have a two variable function and I want to calculate the limit for (x,y) to (0,0) I can just replace x and y with the polar coordinates and check whether it depends on \theta and if it does, there is no limit?
Your missing one small detail given by @pasmith ; take the limit as ##r\to 0##. Then you see if that limit depends on ##\theta##, ...
 
  • #8
WWGD said:
Maybe I shouldn't, but I'd suggest you consider 2 or more random paths of approach. If you get the same limit on both, the limit likely exists. Good if you're stuck in an exam with little time left. Easy choices: along ##y=x## along ##(0,y) ##or ##(x,0)##

Consider [tex]
f(x,y) = \frac{xy}{x^2 + y}.[/tex] Here approaching the origin along a straight line path we find [tex]
\lim_{t \to 0} f(t\cos \theta,t\sin\theta) = \lim_{t\to 0} \frac{t\sin 2\theta}{2(t\cos^2 \theta + \sin \theta)} = 0.[/tex] However, we also find that for [itex]C \neq 0[/itex] we have [tex]
\lim_{t \to 0} f\left(t, \frac{Ct^2}{t-C}\right) = C.[/tex] Hence [itex]\lim_{(x,y)\to(0,0)} f(x,y)[/itex] does not exist: there are points arbitrarily close to the origin where [itex]f(x,y) = C[/itex] for any [itex]C[/itex].
 
  • Like
Likes ananonanunes and SammyS
  • #9
@pasmith , not saying your method doesn't work, just offering a different approach, perspective.
 
  • #10
WWGD said:
@pasmith , not saying your method doesn't work, just offering a different approach, perspective.

My point is that "consider 2 or more random paths of approach. If you get the same limit on both, the limit likely exists" is not accurate.
 
  • #11
pasmith said:
My point is that "consider 2 or more random paths of approach. If you get the same limit on both, the limit likely exists" is not accurate.
" Likely"
It's a rule of thumb. Besides, the limits are different for ## C\neq 0 ##, which doesn't qualify for my rule ( of thumb)
 

What is a multi variable function?

A multi variable function is a mathematical function that has more than one independent variable. This means that the output of the function depends on the values of multiple variables, rather than just one.

How do I find the limit of a multi variable function?

To find the limit of a multi variable function, you must approach the limit point from all possible paths. This involves taking the limit along different paths and comparing the results. If the limit is the same along all paths, then that is the limit of the function. If the limit is different along different paths, then the limit does not exist.

Why is finding the limit of a multi variable function important?

Finding the limit of a multi variable function is important because it can help us understand the behavior of the function as the independent variables approach certain values. It can also help us determine if the function is continuous at a certain point.

What are some common techniques for finding the limit of a multi variable function?

Some common techniques for finding the limit of a multi variable function include using the squeeze theorem, using polar coordinates, and using L'Hopital's rule. These techniques can help simplify the function and make it easier to determine the limit.

Can the limit of a multi variable function be different along different paths?

Yes, the limit of a multi variable function can be different along different paths. This means that the limit may not exist at a certain point, as the function may approach different values depending on the path taken. This is why it is important to approach the limit from all possible paths when finding the limit of a multi variable function.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
787
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
17
Views
575
  • Calculus and Beyond Homework Help
Replies
2
Views
522
  • Calculus and Beyond Homework Help
Replies
8
Views
640
  • Calculus and Beyond Homework Help
Replies
6
Views
825
  • Calculus and Beyond Homework Help
Replies
3
Views
635
  • Calculus and Beyond Homework Help
Replies
14
Views
616
  • Calculus and Beyond Homework Help
Replies
2
Views
110
  • Calculus and Beyond Homework Help
Replies
2
Views
692
Back
Top