# multivariate limits; general method for finding limit…for functions of 2 variables

$\lim_{(x,y)\rightarrow(0,0)}\frac{\sin(x)\sin(y)}{x^2+y^2}$ the limit is undefined.

I would like to know if the following method is sufficient to prove that the limit is undefined:

step1: $$\lim_{(x,y)\rightarrow(x,0)}\frac{\sin(x)\sin(0)}{x^2+0^2}=0$$

step2: $$\lim_{(x,y)\rightarrow(0,y)}\frac{\sin(0)\sin(y)}{0^2+y^2}=0$$

step3: but $$\lim_{(x,y)\rightarrow(x,x)}\frac{\sin(x)\sin(x)}{x^2+x^2}=\frac{\sin(x)^2}{2x^2}=\frac{1}{2}(\frac{\sin x}{x})^2$$

and as $x$ then $\rightarrow 0$ $\frac{\sin(x)^2}{2x^2}=\frac{1}{2}(\frac{\sin x}{x})^2 \rightarrow \frac{1}{2}$

hence the limit does not exist.

1. Does this make sense?
2. Does this method (the '3 steps') always work for functions of two variables?
3. Would it be valid to use the same method to prove: $\lim_{(x,y)\rightarrow(0,0)}\frac{x^3-y^3}{x^2+y^2}=0$

Thank you!

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All steps above look right. –  Babak S. Oct 17 '12 at 6:06
I'm wondering though would the same method work for every 2-variable function; i.e. (1) sub. into limit (x,0) (2) sub. into limit (y,0) and (3) sub. into limit (x,x). If all 3 answers give consistent solutions then the limit exists; if there is at least one inconsistent answer then there is no limit. So my question is would i sometimes need to take another arbitrary step 4) i.e. sub into limit (x,2x); sub into limit (x,x^3) etc... and exhaust every possible step conceivable until there is an inconcistency... –  redrum Oct 17 '12 at 6:12
If they are inconsistent then the limit doesn't exist, but if they are consistent then you haven't proven anything. In fact there are functions that have a consistent limit travelling on any line, but not along non-linear paths, meaning that the limit still doesn't exist. Wikipedia has a good discussion on this here. –  Robert Mastragostino Oct 17 '12 at 6:39
I see, thank you; wikipedia states: "Continuity in each argument does not imply multivariate continuity." This means, my argument may only be used to prove there is no limit. Is there any other method in which I may prove there is a limit, i.e. for my second example in "step 3." –  redrum Oct 17 '12 at 6:48
I have my answer for the example I solved above; but for the other example below: lim(x,y)--> (0,0) for {x^3-y^3}/{x^2+y^2}=0; I have no idea how to prove mathematically. –  redrum Oct 17 '12 at 8:36

About the third: In this kinds of limits, you can take $r_α (t)=(t,αt)$ and put it into your function to find path wise limit of $f$ when $t$ tends to $0$ . After simplifying the original function, if the limit of last expression approaches to zero then probably your original function has limit $0$ at $(0,0)$ . Now, you have to use $ϵ,δ$ to prove your limit. Note that we see $\lim_ {t→0} r_α (t)=(0,0)$ . There are some small magic points in these limits also. See what @Jennifer Dylan here noted http://math.stackexchange.com/a/128564/8581. Read two comments below of it and take a great way for such these limits. I hope it helps you.