$0/0$ limit question 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?
 A: 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?).
A: 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) = (t, mt), 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) = (t, mt), we just take T = 0, and evaluate the limit of f(t, mt) as t approaches 0. Up to a change of variables, this is just what you have done above.
A: 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 $x\to0^+$.
A: 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 $(*)$.
A: 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,
$$\lim_{(x,y) \to (0,0)} f(x,y)=0$$.
A: 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.
