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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.2010-08-18
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    @Chandru1: Yes, you're right. Thanks for the correction. It says that "y is a straight line with any value for m but -1"2010-08-18

6 Answers 6

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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?2010-08-18
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    0 + 0m = 0 so limit = 0/0, so they are using L'hospital's rule.2010-08-18
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    deinst, no one is using L'Hopital's rule. Kokoloko, would you have the same "problem" with $\lim_{x\to0}x/x$?2010-08-18
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    @Kokoloko: You could just cancel the x on top and bottom.2010-08-18
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    @Robin Chapman: Yes. What is it that I'm missing?2010-08-18
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    @Niel de Beaudrap: So true. I can't believe I didn't see that. So that part is explained.2010-08-18
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    **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: http://groups.google.com/group/sci.math/msg/465b5953538f40492010-09-29
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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?2010-08-18
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    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.2010-08-18
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    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.2010-08-18
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    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. :)2010-08-18
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    @Robin :D And how can I find out the answer?2010-08-18
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    Kokoloko: Have you tried plotting the functions you're trying to take limits of? It might help you in understanding why the limits behave the way they do.2010-08-18
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    @J. Mangaldan: I've tried, but I can't isolate the "y", so I've only got a limited drawing with the numbers I've obtained.2010-08-18
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    I mean a surface plot. You can use something like http://www.fooplot.com/index3d.php2010-08-18
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    Ok, I see it now. If I understand it right, "x" and "y" never reach "0" in that surface, so the limit when "x" and "y" tend to "0" doesn't exist. Is that the correct reasoning?2010-08-18
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    Kokoloko, did you take my hint?2010-08-18
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    Robin, we should probably be asking Kokoloko if he already understands the reasoning behind switching to polar coordinates...2010-08-18
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    @Robin I don't think so. I guess you refer to "the answer seems to be 0". I understand that if you say that it should mean that the answer isn't 0, but I'm not sure if there's anything else that you're trying to make me understand there :/. I'm very sorry for making such a hassle out of it, but it's taking me some effort to understand it all.2010-08-18
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    Mmmm... I've tried passing the function to polar coordinates and I get: f(r,o)=-rcos^3(o)+3rcos(o)sin^2(o) . Is that right? How should I transform the limit tendency now? (can you please confirm if the reasoning above last Robin's comment is right? It would help me know if I'm starting to understand something :) )2010-08-18
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    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 $r(-\cos^3 t+3\cos t\sin^2 t)$?2010-08-18
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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.2010-08-18
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    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).2010-08-18
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    @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.2010-10-20
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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.2010-08-18
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    E.g. choose $x^{1/2},\; cx,\; x^2\;$ for $f(x)$2010-08-18
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    Thanks! Let me try that ;)2010-08-18
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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!2010-08-18
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    L'Hopital's rule should be avoided wherever possible. Using it here is just ludicrous.2010-08-18
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    @Robin: Why???? (Yes, one question mark would do, but not for stackexchange, which insists on at least 15 chars for a comment)2010-08-18
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    Search the archives of sci.math for my collected rants about L'H's rule :-)2010-08-18