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I have an examples book with a limit exercise that I can't understand.

The limit in question is:

$$f(x,y)=\frac{x}{x+y}$$ with $x\ne-y$; $$\lim_{(x,y)\to(0,0)} f(x,y)$$

And then to solve it, it goes:

$$\lim_{(x,y)\to(0,0)} f(x,y) = \lim_{x\to0} f(x,mx) =\lim_{x\to 0}\frac{x}{x+mx}=\frac{1}{1+m}.$$

Can you help me understand that? Thanks,

UPDATE: Ok, just to make sure that I got it right. I have a very similar test exercise with $4$ different options.

The following limit $$\lim_{(x,y)\to(0,0)}\frac{-x^3+3xy^2}{x^2+y^2}$$ equals:

A. $0$

B. $- \infty$

C. Doesn't exist

D. $ +\infty$

My doubt is: if I consider it normally I'd say that it doesn't exist, but if I solve it using the same approach (i.e. $y=mx$) then the limit equals $0$. Which one is the right answer?

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I think you are missing something important regarding $y$. Is $y$ a straight line massing through the origin..or anything like this has been mentioned in the book. – anonymous Aug 18 '10 at 11:05
@Chandru1: Yes, you're right. Thanks for the correction. It says that "y is a straight line with any value for m but -1" – Kokoloko Aug 18 '10 at 11:50

The point is that this limit doesn't exist.

For it to exist, then if you take any sequence of points $(x_n,y_n)$ converging to $(0,0)$ for which $f(x_n,y_n)$ is defined, then $$\lim_{n\to\infty}f(x_n,y_n)$$ must exist and also be independent of the sequence $(x_n,y_n)$. The book is saying in essence that if you have a sequence of points of the form $(x_n,m x_n)$ (so lying on the line with gradient $m$ though the origin) and converging to the origin, then $$\lim_{n\to\infty}f(x_n,y_n)=\frac1{1+m}\qquad\qquad\qquad(*)$$ which is not independent of the sequence of points $(x_n,y_n)$, as changing $m$ will change the limit $(*)$.

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Your answer is very accurate and similar to the one I have in the book. The problem I have, and I probably didn't explain it well enough is with the last step, where it says that (x->0)lim(x/(x+mx))=1/(1+m). Why isn't it =(0/(0+0m)) thus 0? – Kokoloko Aug 18 '10 at 11:58
0 + 0m = 0 so limit = 0/0, so they are using L'hospital's rule. – deinst Aug 18 '10 at 12:06
deinst, no one is using L'Hopital's rule. Kokoloko, would you have the same "problem" with $\lim_{x\to0}x/x$? – Robin Chapman Aug 18 '10 at 12:08
@Kokoloko: You could just cancel the x on top and bottom. – Niel de Beaudrap Aug 18 '10 at 12:09
REMARK While it's true that L'Hospital's rule is overkill here, as I remarked before on sci.math, it's a good thing that the rule applies in such cases because in general there is no algorithm for deciding if two functions are equal (or 0), so e.g. one cannot tell in general if f = 0 or f = g in f/g, but the rule may still apply and yield the correct result (whether by human or machine computation). See this thread for further discussion: – Bill Dubuque Sep 29 '10 at 16:50

For your second question: is $$\lim_{(x,y)\to(0,0)}\frac{-x^3+3xy^2}{x^2+y^2}=0?$$ imagine a sequence of points $(x_n,y_n)$ converging to the origin, and think whether $f(x_n,y_n)\to0$.

I would be tempted to write these points in polar form: $x_n=r_n\cos t_n$ and $y_n=r_n\sin t_n$ as then the denominator becomes $r_n^2$ (nice) and moreover $r_n\to0$ (why?).

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Mm... I'm not sure if I did it right, but giving values close to (0,0) to the variables seems to get results close to 0. Does that mean that the answer is 0 then? – Kokoloko Aug 18 '10 at 13:22
Now that I think about it, changing the Cartesian variables in the original question to polar form also makes it clear that the limit does not exist: the role played here by $\theta$ is the same role played by m in your first answer. – J. M. Aug 18 '10 at 13:30
Kokoloko, what it means is that the answer seems to be $0$ :-) (whether it is zero is another matter). Jan, changing to polars is generally a good trick when dealing with homogeneous functions. – Robin Chapman Aug 18 '10 at 14:30
Oh, you don't have to worry about me not appreciating polar/cylindrical/spherical transformations, I've seen them vastly simplify operations that are otherwise hard to do the Cartesian way. :) – J. M. Aug 18 '10 at 14:40
Good: so you get $r(-\cos^3 t+3\cos t\sin^2 t)$. As $(x_n, y_n)\to0$ then $r_n\to0$. So if $r$ tends to zero, what can you say about $r(-\cos^3 t+3\cos t\sin^2 t)$? – Robin Chapman Aug 18 '10 at 18:18

I will elaborate on Robin's answer.

There is no unique limit as (x,y) goes to (0,0) --- i.e. the limit does not exist. For instance, if we approach (0,0) on the x axis, we obtain the limit

$\lim\limits_{(x,y) \to (0,0)} \dfrac{x}{x+y} = \lim\limits_{x \to 0} \dfrac{x}{x} = 1,$

and if we approach (0,0) on the y axis, we obtain the limit

$\lim\limits_{(x,y) \to (0,0)} \dfrac{x}{x+y} = \lim\limits_{y \to 0} \dfrac{0}{y} = 0.$

However, we can still consider limits along different curves. The two above limits, along the x and y axes, are two examples: the limit you describe in your question is a similar limit, on the locus of the equation y = mx. This is something different than whether there is "a limit" of f(x,y) as we approach (0,0) --- in this case, because there are different limits depending on how we approach the origin, we may consider the question of what limit one has as one approaches the origin on a particular curve.

In this case, a "curve" is a function

$\big(x(t),y(t)\big) = c(t)$,

for some function $c: \mathbb R \to \mathbb R^2$. "Approaching" (0,0) along the curve c(t) entails specifying some domain for c in which it does not cross itself (for the example of the line y = mx one could take c(t) = (tmt), in which case the domain can be the real numbers) and for which c(t) = (0,0) for some (unique) value t = T. Then, evaluating the limit of f(x,y) along the curve (x,y) = c(t) means just taking the limit of the composite function f(c(t)) as t approaches T.

In the case of approaching (0,0) on the curve c(t) = (tmt), we just take T = 0, and evaluate the limit of f(tmt) as t approaches 0. Up to a change of variables, this is just what you have done above.

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Your answer does a great job elaborating on Robin's answer. I think I'm getting the grasp of it. I understand the your examples of approaching the x and y axis, but I don't get the one of my book (with straight lines). If you could elaborate on that I'd really appreciate it. – Kokoloko Aug 18 '10 at 12:26
Is there something specific that you don't understand in my lower two paragraphs? Think of x and y as positions of some particle, and then let these vary with a "time" variable. The limit along a curve (or line) is the limit of f(x(t),y(t)), along that trajectory. The X and Y axes are just two possible trajectories, in which the y and x co-ordinates (resp.) are constant with time; other lines are a more general class of the same sort of thing, and special cases of the class of all continuous trajectories (x(t), y(t)) = c(t). – Niel de Beaudrap Aug 18 '10 at 13:17
@Kokoloko Niel's answer is correct to the problem as posted. In the comments to your question you say the limit is taken along a line through the origin. This is what permits the book to substitute mx for y. Normally (and in your second example) it is required that the function be close to the limit for all points close to the target. If you take different paths approaching the target and get different limits, the limit doesn't exist. – Ross Millikan Oct 20 '10 at 22:27

HINT The limit becomes that of $x/f(x)$ along the path $y = f(x)-x$ where $f(x)\to 0$. But this limit can be whatever you desire, by choosing $f$ weaker, comparable, or stronger than $x$ at $0$. Explicitly: choosing $f$ to be one of $\; x^{1/2},\; x/c,\; x^2$ yields $\; x/f(x)\to 0,\; c,\; \infty\;$ respectively, as $x\to0^+$.

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I'm sorry, but I don't understand your hint. – Kokoloko Aug 18 '10 at 15:46
E.g. choose $x^{1/2},\; cx,\; x^2\;$ for $f(x)$ – Bill Dubuque Aug 18 '10 at 16:00
Thanks! Let me try that ;) – Kokoloko Aug 18 '10 at 16:16

EDITED. The version I originally posted was wrong.

For the second problem, use the fact that squares are non-negative to show that $$|f(x,y)| \le 3|x|$$

From which it follows (by using the sandwich theorem) that, $$\lim_{(x,y) \to (0,0)} f(x,y)=0$$.

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De l'Hôpital explains the last step.
If both numerator and denominator approach zero, you can apply de l'Hôpital, provided the derivative of the functions in numerator and denominator exist for x=0, which is the case.

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That is way more than overkill... x is not zero, so we can cancel it from the top and bottom in the last step! – Dylan Wilson Aug 18 '10 at 12:45
L'Hopital's rule should be avoided wherever possible. Using it here is just ludicrous. – Robin Chapman Aug 18 '10 at 13:01
@Robin: Why???? (Yes, one question mark would do, but not for stackexchange, which insists on at least 15 chars for a comment) – stevenvh Aug 18 '10 at 13:11
Search the archives of sci.math for my collected rants about L'H's rule :-) – Robin Chapman Aug 18 '10 at 14:31

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