This is a relatively strange question, title says the original question. Intuition says this limit is 1, and graphing the function $f(x,y)=\frac{x}{y}$ this confirms it.
$$\lim_{x \rightarrow \infty}(\lim_{y \rightarrow \infty} (\frac{x}{y}) ) = 1$$
I do not know how, but online calculators disagree. Wolfram alpha widgets seems to think that this limit doesn't exist and symbolab somehow ended up with 0.
However this is where I don't understand certain things. I encountered this issue in the context of converting an infinite sum to an integral.
It is known that an infinite sum can be often rewritten as a Reimann sum in the following way:
$$\int_{a}^{b}f(x)dx=\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(f(a+\frac{b-a}{n}k))\frac{b-a}{n}$$
Various numerical examples suggest that this is true. The next step is where I do not understand where I went wrong. It makes sense to think that if we are taking an integral from $0$ to $\infty$ we can write it in the following way:
$$\lim_{b \rightarrow \infty}\int_{0}^{b}f(x)dx=\lim_{b \rightarrow \infty}\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(f(\frac{b}{n}k))\frac{b}{n}$$
Since both limits go to infinity, the quantity of $\frac{b}{n}$ should go to one, as previously established. So I expected the following to hold:
$$\lim_{b \rightarrow \infty}\int_{0}^{b}f(x)dx=\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(f(k))$$
However, this apparently isn't true, according to numerical calculations. Consider the following infinite sum, stolen shamelessly from someone else's question:
$$\sum_{k=1}^{\infty}(\frac{1}{5^k+2})$$
Letting $f(x)=\frac{1}{5^k+2}$ we have:
$$\sum_{k=1}^{\infty}(\frac{1}{5^k+2})=\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(f(k))=\lim_{b \rightarrow \infty}\int_{0}^{b}f(x)dx=\lim_{b \rightarrow \infty}\int_{0}^{b}\frac{1}{5^x+2}dx$$
$$\sum_{k=1}^{\infty}(\frac{1}{5^k+2})=\int_{0}^{\infty}\frac{1}{5^x+2}dx$$
According to desmos, the RHS evaluates to ~0.3413 and the LHS evaluates to ~0.1898. So this begs the question of what I did wrong. For anyone interested, here are the images of $\int_{0}^{x}\frac{1}{5^{t}+2}dt$ and $\sum_{n=1}^{x}\left(\frac{1}{5^{n}+2}\right)$ in red and blue respectively, drawn by desmos.
Best Answer
The problem you're running into is that limits are much more slippery in multiple variables. You have this two variable function $f(x,y)=x/y$. Somehow you want to ask "what does this function do as $x$ and $y$ both go to infinity?" The problem is that there are many ways to go off to infinity. For instance, we could march off along the line $y=x$, which seems to be what you've done. Restricting to this line, we have $$\lim_{x \to \infty} f(x,x) = \lim_{x \to \infty} 1 = 1.$$ But you could've gone off along the line $y=2x$ and gotten a limit $$\lim_{x \to \infty} f(x,2x) = \lim_{x \to \infty} 1/2 = 1/2.$$ Even worse, you could've gone off along a parabola $x=y^2$ and seen $$\lim_{y \to \infty} f(y^2,y) = \lim_{y \to \infty} y = \infty.$$
Another approach is that the limit $\lim_{x \to \infty, y \to \infty} f(x,y)$ should be the same as $\lim_{(x,y) \to (0,0)} f(1/x,1/y)$. But in your case $f(1/x,1/y)= y/x$, and this function clearly doesn't have a limit at the origin (nor is it defined there).