# Why does the limit not exist?

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1. Sep 28, 2016

### toforfiltum

1. The problem statement, all variables and given/known data
This is the function:

$\lim_{(x,y) \rightarrow (0,0)} \frac{(x+y)^2}{x^2+y^2}$

2. Relevant equations

3. The attempt at a solution
So for $x \rightarrow 0$ along $y=0$, $f(x,y)=1$

For $y \rightarrow 0$ along $x=0$, $f(x,y)=1$ also.

But the answer says there is no limit that exists.

Is it because I didn't try approaching (0,0) using other functions?

So, if I try it for $y=mx \rightarrow 0$,

$f(x,y) = \frac {(x+mx)^2)}{x^2+m^2x^2}$

$=\frac {(1+m)^2}{1+m^2}$, which is $\neq 1$ since $m \neq 0$

Is this the reason why? And if so, what methods should I use to evaluate if functions truly have a limit or not? What if for another case, the above three cases gave a same answer, but there is another function, say $y=x^2$ which will show that the function actually doesn't have a limit? How do I actually know for sure if a function has a limit or not?

2. Sep 28, 2016

### Lucas SV

Consider approaching $(0,0)$ along the curve given by $y=-x$.

3. Sep 28, 2016

### toforfiltum

Ah yes, I see what you mean. I would get zero instead. Thanks.

4. Sep 28, 2016

### Ray Vickson

You have already answered your own question! IF a limit did exist you would have to get the same result for any line $y = kx$ or $x = m y$ or, more generally, for any smooth curves $y = g(x)$ or $x = h(y)$ that passed through $(0,0)$.

In a case like this one, it helps to go to polar coordinate $x = r \cos(\theta), \: y = r \sin(\theta)$. For $(x,y) \neq (0,0)$ (that is, $r \neq 0$) that gives you
$$f(x,y) = \frac{(r \cos(\theta) + r \sin(\theta))^2}{r^2} = (\cos(\theta)+ \sin(\theta))^2 = 1 + 2 \sin(\theta) \cos(\theta).$$

5. Sep 28, 2016

### toforfiltum

So when converted to polar coordinates, $lim_{r \rightarrow 0} f(r \cos\theta, r\sin\theta) = 1 + 2 \sin(\theta) \cos(\theta)$.

I can prove that the limit does not exist for this function because I get different answers when I input different values of $\theta$, right?

6. Sep 28, 2016

### Ray Vickson

You tell me.

7. Sep 28, 2016

### toforfiltum

I don't really understand what you mean but yes.

8. Sep 28, 2016

### Lucas SV

There is a simple theorem which states that if the limit exists, it must be unique. There is another theorem which states that if the limit exist, it must be equal to the limit from any direction. So you assume there is a limit, you will get a contradiction.

9. Sep 28, 2016

### toforfiltum

So basically what I'm saying about the function having different outputs with different inputs of $\theta$ is the same as the theorem, right?

10. Sep 28, 2016

### Ray Vickson

Good, that was the answer I wanted.

Last edited: Sep 28, 2016
11. Sep 28, 2016

### toforfiltum

Nice.

12. Sep 28, 2016

### Lucas SV

Yes, it is the same as the second theorem I mentioned (i edited just after your reply).

13. Sep 28, 2016

### toforfiltum

Thanks!

14. Sep 28, 2016

### LCKurtz

I don't think anyone has answered this part of your question. You do understand that if you get different answers along different paths there is no limit. That's fine. But, like you have observed, what if all your your choices of paths and keep getting the same number? Then you have to wonder, maybe the limit really does exist, but this doesn't prove it, or maybe the limit doesn't exist and I just haven't been clever enough choosing different paths. That is a very reasonable question to ask.

Most first year calculus books will give problems where the limit doesn't exist and it isn't too difficult to find paths that prove it, and maybe a problem or two where the limit does exist. The students usually find this second type to be difficult. That's because they are difficult because they may require methods you haven't seen or thought of. Don't worry too much about that. This topic is covered in more detail in advanced calculus courses where you really get into $\delta -\epsilon$ proofs and limit theorems. So I guess the short answer is: don't lose too much sleep over this right now; wait until you take advanced calculus and you can lose sleep then

15. Sep 28, 2016

### toforfiltum

If so, I should definitely lose sleep now. I don't know how advanced multivariable calculus is, but that's the course I'm taking right now, and the problems I have posted so far come from Susan Colley's Vector Calculus textbook. My professor has also covered the $\epsilon-\delta$ proofs and limit theorems, though I must profess that I don't really get it. As it is, I'm losing much sleep over this course now.

I know in my gut that I"m not smart enough to be a math major, but I would like to ask your opinion on if it's important to understand the proofs like these? I'm going to be an engineer, so I would probably just use math as a tool right? I've tried to understand the proofs, but at times, they are too abstract so they seem not intuitive to me.

And seeing that I've gotten to this point, what's the answer to the question I posted?

16. Sep 28, 2016

### LCKurtz

Of course it's good to understand as much as possible. Some things take longer to "soak in". You may encounter material that you don't quite understand now but will understand a year later after you have had more experience. I know because that happened to me when I hit advanced calculus. But, concerning the current topic, most functions you will ever encounter are continuous most everywhere and the places where thing go wrong are where you have zeroes in denominators or negative numbers in square roots and the like. Math textbooks are probably the only place you will encounter this type of problem.

So, no, I don't think the fact that you are having trouble with this topic at this time will have any effect on your career as an engineer. Stick with it and learn as much as you can. You will find that even the course you are in will soon move to other topics you may like better. The analytical thinking you learn in math courses is what will help you as an engineer, even if you never see some of the specific topics again.

17. Sep 28, 2016

### toforfiltum

You have no idea how comforting your words are. Right now, I'm putting in so many hours for this course just to stay afloat. Lectures cover so many topics at such a fast pace, I'm frightened. So thank you very much for your kind words, and I'll continue seeking help here.

18. Sep 29, 2016

### Ray Vickson

BTW: that is exactly WHY I suggested looking at polar coordinate!.

To have a well-defined limit $c$ as $(x,y) \to (0,0)$ we want $f(x,y)$ to be going to $c$ for $(x,y)$ nearer and nearer $(0,0)$. The nice thing about polar coordinates---in this type of problem, at least---is that the single variable $r$ measures the distance to $(0,0)$, so if for any $\theta$ you have $f(r,\theta)$ going to the same value $c$ as $r \to 0$, you have a limit, and it equals $c$.

For most problems in first or second calculus courses, polar coordinates can clear up the majority of the problems that are left when cartesian coordinates $(x,y)$ product difficulties. Of course, there will always be some really nasty problems where even polar coordinates fail to help, but it is unlikely you would encounter those on important events such as exams, etc.

Last edited: Sep 29, 2016
19. Sep 29, 2016

### epenguin

You had it in your hands in your formula (1 + m)2/(1 + m2) of #1 which takes every value between 0 amd +∞ !

20. Sep 29, 2016

### toforfiltum

Hmm...I have a minor confusion here. Is it possible for a function $f(r,\theta)$ to go to a value of $c$ when $r \rightarrow 0$? I've always thought that since $(x,y) \rightarrow (0,0)$ then $r \rightarrow 0$ so limit of $f(r, \theta)$ also should tend to zero, if there's a limit? Could you provide an example that says otherwise?

And thanks for the tip! Would definitely keep that in mind when I'm taking my midterm next week!