Limit question as $x$ and $y$ approach infinity? I have to prove that the limit of the function $\frac{x^2}{x^2+y^2}$ as $x$ approaches infinity and as y approaches infinity does not exist.
I thought about finding the side limits, and if they are not equal, bam! I have solved it. But what should I take as side limits here? $+$ and $-$ infinity? Thank you in advance :)
 A: $$(1)\;\;\;\;y=x\;\;\;\;\Longrightarrow\;\;\; \lim_{x,y\to\infty}\frac{x^2}{x^2+y^2}=\lim_{x\to\infty}\frac{x^2}{2x^2}=\frac{1}{2}$$
$$(2)\;\;\;\;\;\;\;\;\;\;\;\;\;y=x^2\;\;\;\;\Longrightarrow\;\;\; \lim_{x\to\infty}\frac{x^2}{x^2+x^4}=0$$
A: ok at first we say we observe $x\to \infty$ and $x=y$ than the limit is the same as
$$\lim_{x\to \infty}\frac{x^2}{2x^2}=\frac{1}{2}$$
Now we look at $x=2y$ than the limit is 
$$\lim_{y \to \infty} \frac{4y^2}{4y^2 +y^2}=\frac{4}{5}$$ but the limits must be same so they don't exists.
A: Let $x=r\cos\theta$ and $y=r\sin\theta$. Then
$$\frac{x^2}{x^2+y^2}=\cos^2\theta$$
if $(x,y)\ne (0,0)$. Thus even for very large positive $x$ and $y$, the function $\frac{x^2}{x^2+y^2}$ can take on any value in the interval $(0,1)$. 
A: The point you need to know, and glean from excellent examples posted, is to check the limit of your function as $(x, y) \to \infty$ along two or more different curves. 
You are working in two dimensions, so the choice of curves you can test can is not one dimensional lines. Put, say, $y = x$, and $y = x^2$, perhaps even $y = x^3$. 
Then, write the function as a function of $x$ (replace $y$ in the function $f(x)$ for each curve, depending on the curve $y$) and evaluate limit of the function as $x \to \infty$. 
E.g., with your function, if we let $y = x$, then substitute $x$ whenever $y$ appears in you function, and evaluate: $$\lim_{x\to \infty} \frac{x^2}{2x^2} = 1/2\tag{1}$$
And if $y = x^{-2}$, we substitute $x^{-2}$ for $y$ and then evaluate $$\lim_{x\to \infty} \frac{x^2}{1 + x^2} = 1\tag{2}$$
If the limits as $x\to \infty$ of the functions defined in terms of different curves, are not equal, as we have with $(1), (2)$, you know the limit does not exist.
A: I like Andre Nicolas' answer and mine is similar. Since the degree of the terms in the denominator are of equal degree, let $$y=ax.$$
Then we have $$1/(1+a^2).$$
Suppose you have $$x^2 +y^4$$ in the denominator, with $$y^4$$ in the numerator. Then let $$x=ay^2.$$
So you see that this approach is generalizable.
