Thread: limit of x/x as (x,y)->(0,0)

1. limit of x/x as (x,y)->(0,0)

lim x/x
(x,y)->(0,0)

i am a little confused about multivariate limits. i know that for a single variable, this limit is trivial and is simply 1, but for this limit i suspect it to be 1 but i am not sure. based on the definition of the limit, it says that for all epsilon there exists a delta such that when the distance between (x,y) and (0,0) is less than delta, then the distance between f(x) and the limit L is less than epsilon. but this function x/x is not defined along the line x=0. this means that in some delta neighborhood of (0,0) the function will not be defined there.

some definitions of limits i read say that x must be in the domain of the function, but other definitions don't seem to say it explicitly.

for example (from wikipedia):
Let ƒ be a function defined on an [COLOR=rgb(0, 0, 0)]open interval[/COLOR] containing c (except possibly at c) and let L be a real number. then for each real ε > 0 there exists a real δ > 0 such that for all x with 0 < |xc| < δ, we have |ƒ(x) − L| < ε

this definition does not seem to mention that fact that x must be in the domain of the function. is this definition incorrect then?

2. Originally Posted by oblixps
lim x/x
(x,y)->(0,0)

i am a little confused about multivariate limits. i know that for a single variable, this limit is trivial and is simply 1, but for this limit i suspect it to be 1 but i am not sure. based on the definition of the limit, it says that for all epsilon there exists a delta such that when the distance between (x,y) and (0,0) is less than delta, then the distance between f(x) and the limit L is less than epsilon. but this function x/x is not defined along the line x=0. this means that in some delta neighborhood of (0,0) the function will not be defined there.

some definitions of limits i read say that x must be in the domain of the function, but other definitions don't seem to say it explicitly.

for example (from wikipedia):
Let ƒ be a function defined on an [COLOR=rgb(0, 0, 0)]open interval[/COLOR] containing c (except possibly at c) and let L be a real number. then for each real ε > 0 there exists a real δ > 0 such that for all x with 0 < |xc| < δ, we have |ƒ(x) − L| < ε

this definition does not seem to mention that fact that x must be in the domain of the function. is this definition incorrect then?
I don't understand the confusion. In general $\displaystyle \lim_{x\to c}f(x)$ is intelligible even if $c\notin\text{Dom }f$. Right? Take for example this limit in the one-dimensional case. The rule is that $\displaystyle \lim_{x\to c}f(x)=L$ then for every $\varepsilon>0$ there exists some $\delta>0$ such that $0. Take $\varepsilon=\delta$ in your case, no?

3. Originally Posted by oblixps
lim x/x
(x,y)->(0,0)

i am a little confused about multivariate limits. i know that for a single variable, this limit is trivial and is simply 1, but for this limit i suspect it to be 1 but i am not sure. based on the definition of the limit, it says that for all epsilon there exists a delta such that when the distance between (x,y) and (0,0) is less than delta, then the distance between f(x) and the limit L is less than epsilon. but this function x/x is not defined along the line x=0. this means that in some delta neighborhood of (0,0) the function will not be defined there.

some definitions of limits i read say that x must be in the domain of the function, but other definitions don't seem to say it explicitly.

for example (from wikipedia):
Let ƒ be a function defined on an [COLOR=rgb(0, 0, 0)]open interval[/COLOR] containing c (except possibly at c) and let L be a real number. then for each real ε > 0 there exists a real δ > 0 such that for all x with 0 < |xc| < δ, we have |ƒ(x) − L| < ε

this definition does not seem to mention that fact that x must be in the domain of the function. is this definition incorrect then?
Did you right the question properly? Since the expression you gave is independent of y, $\lim\limits_{(x,y)\to(0,0)}\dfrac{x}{x}=\lim\limit s_{x\to0}\dfrac{x}{x}=1$...

However, I think you meant to say $\lim\limits_{(x,y)\to(0,0)}\dfrac{x}{y}$ or $\lim\limits_{(x,y)\to(0,0)}\dfrac{y}{x}$

4. no i wrote the question correct. the function is x/x which just cancels and equals 1. what i was confused about was that in the definition of the limit, whenever x is within delta of 0, then f(x) is within epsilon of the limit. but this function is undefined along the like x=0 as you can see. in the wikipedia definition of the limit it does not require that the x's have to be in the domain of the function. or is this something just implied by the definition? in my book they explicitly state that the x's must be in the domain of the function f but on wikipedia and other sites i have visited, their definition of the limit does not include that explicit statement. this is the source of my confusion.

5. To find $\displaystyle \lim_{x \to c}f(x)$, $\displaystyle f(x)$ need not be continuous at $\displaystyle x = c$, because you are determining the behaviour NEAR that point, not AT that point.

6. yes so x/x is not defined at the point (0,0). but x/x isn't even defined along the line or in this case plane x=0. it is not just 1 point where the function does not exist. it is many points, infinite in number in fact. i know that a function does not need to be defined at a point for the limit to exist there but with the function not being defined on a whole plane of points, would that pose a problem?

7. Oops! All of this confusion is making me confused.

Post removed.

8. Originally Posted by oblixps
lim x/x
(x,y)->(0,0)

i am a little confused about multivariate limits. i know that for a single variable, this limit is trivial and is simply 1, but for this limit i suspect it to be 1 but i am not sure. based on the definition of the limit, it says that for all epsilon there exists a delta such that when the distance between (x,y) and (0,0) is less than delta, then the distance between f(x) and the limit L is less than epsilon. but this function x/x is not defined along the line x=0. this means that in some delta neighborhood of (0,0) the function will not be defined there.
I think that you are right to be suspicious of this sneaky question. In order for $\displaystyle\lim_{(x,y)\to(0,0)}f(x,y)$ to exist, it is not necessary for $f(0,0)$ to be defined. But (unless indicated otherwise) the definition does require that the function should be defined in some punctured neighbourhood of the limit point. In this case, as you correctly say, the domain of the function does not include the y-axis, and therefore does not include any punctured neighbourhood of (0,0). So strictly speaking one ought to say that the limit is not defined.

9. Deleting to avoid confusion

10. The formula $f(x,y)=x/x$ defines the function:

$f:S\subset \mathbb{R}^2 \rightarrow \mathbb{R},\;\;S=\left\{{(x,y)\in \mathbb{R}^2:x\neq 0}\right\}$

$f(x,y)=1$

This means that $(0,0)$ is an accumulation point of $S$ and as a consequence it has sense to ask if there exists

$L=\displaystyle\lim_{ (x,y) \to (0,0)}f(x,y)$

In this case, and choosing $\delta=\epsilon$ (as Drexel28 did) we conclude that the limit is $L=1$ .

Fernando Revilla

11. Originally Posted by DrSteve
Deleting to avoid confusion
I'm sorry that this comment was deleted, because I thought it made a very good point:
Originally Posted by DrSteve (deleted)
An example analogous to this in one-variable calculus would be $\lim_{x\rightarrow 1^+}\sqrt{1-x^2}$ which doesn't exist simply because the function isn't defined to the right of 1 (note that the graph of the function is a semicircle of radius 1).
That is a good analogy. It's okay to say that $\lim_{x\rightarrow 1^-}\sqrt{1-x^2} = 0$, because the limit from the left is zero.

But it's not good to write about $\lim_{x\rightarrow 1}\sqrt{1-x^2}$, because the (two-sided) limit doesn't exist.

12. Originally Posted by Opalg
But it's not good to write about $\lim_{x\rightarrow 1}\sqrt{1-x^2}$, because the (two-sided) limit doesn't exist.
Not good for an analyst, but good for a topologist.

Fernando Revilla

13. FernandoRevilla, i have not learned about accumulation points yet but i checked the definition which says if every neighborhood of a point x as a point other than x than it is an accumulation point. so i can see that (0,0) is an accumulation point. i know that the accumulation point is supposed to be the generalization of the notion of a limit in topological spaces but when you said "This means that (0,0) is an accumulation point of S and as a consequence it has sense to ask if there exists lim (as (x,y)->(0,0)) of f(x,y)" i don't understand. why is it that when (0,0) is an accumulation point of S its possible to talk about the limit? whats the connection between having an accumulation point in the domain and having the limit of the function f(x,y) as (x,y) approaches that accumulation point exist?

14. Originally Posted by oblixps
... so i can see that (0,0) is an accumulation point. i know that the accumulation point is supposed to be the generalization of the notion of a limit in topological spaces
Right. More specific, suppose it in metric spaces. I used that generalization.

but when you said "This means that (0,0) is an accumulation point of S and as a consequence it has sense to ask if there exists lim (as (x,y)->(0,0)) of f(x,y)" i don't understand. why is it that when (0,0) is an accumulation point of S its possible to talk about the limit? whats the connection between having an accumulation point in the domain and having the limit of the function f(x,y) as (x,y) approaches that accumulation point exist?
If a point $a$ is an accumulation point of the domain $S$ of a function $f$ we can assure that $f$ is defined in points near to $a$ just as we want. In our case and using that generalization

$\displaystyle\lim_{(x,y) \to (0,0)}{f(x,y)}=1$

because

$\forall{} \epsilon>0\exists{} \delta>0:\;( 0<\left\|{(x,y)-(0,0)}\right\|<\delta\;\wedge\;(x,y)\in S)\Rightarrow \left |{f(x,y)-1}\right |<\epsilon$

Note that for every $\delta>0$ we can assure that there exists at least one $(x,y)$ such that the distance of $(x,y)$ to $(0,0)$ is less than $\delta$ and $f(x,y)$ exists.

On the other hand, if you use a more restrictive conditions (for example if the definition does require that the function should be defined in some punctured neighbourhood of the limit point (see Opalg's post) then clearly the limit does not exist.

As you see, this depends on the context. Ignoring that context, I used the most generalizated definition considering $\mathbb{R}^2$ and $\mathbb{R}$ as metric spaces with the usual distances.

Fernando Revilla

15. It's been a while since I've done Multivariate Analysis.

Fernando, are you saying that there is more than one variation on the definition of a limit? I originally thought that the definition Opalg was using was correct, but I was then swayed by other arguments in this thread (which is why I deleted my post).

Does the "correct" definition just depend on personal taste and/or the author of the textbook you're using. Or are there different definitions based on the subject matter you're studying? (Metric Spaces, Real Analysis, Topology, etc.)

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