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. $$\lim_{x \rightarrow \infty}(\lim_{y \rightarrow \infty} (\frac{x}{y}) ) = 1$$

image of graph

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:


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.

image of graph

  • 4
    $\begingroup$ For any $x>0$, $\lim_{y\to\infty}\frac xy=0$. And $\lim_{x\to \infty}0=0$ The case is closed. $\endgroup$
    – Mark Viola
    Nov 6, 2020 at 4:23
  • $\begingroup$ Responding to Viola's comment, let's say your are correct, then does that mean that $\lim_{b \rightarrow \infty}\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(f(\frac{b}{n}k))\frac{b}{n}$ should evaluate to zero? Also consider evaluating the limit along the line $x=y$ maybe? $\endgroup$
    – person
    Nov 6, 2020 at 4:29
  • $\begingroup$ No because you may have nonzero terms in the beginning from when $n$ is small , but eventually your sums tend to zero. Then you take the limit of that as $b$ approaches infinity . $\endgroup$
    – Derek Luna
    Nov 6, 2020 at 4:36
  • $\begingroup$ Luna, if you think you're onto something, please try to formalize it and write it as an answer. I am not looking for a simple comment, but a formalized response. I would be happy to hear what your have to say. $\endgroup$
    – person
    Nov 6, 2020 at 4:39
  • $\begingroup$ First, do not refer to me by my last name! That is extraordinarily rude. My name is Mark. Is that clear? Second, as you wrote the object, $x$ and $y$ are independent and the inner limit is zero. You could write a different limit such as $\lim_{x\to \infty\\y\to\infty}\frac{x}{y}$. Of course $\lim_{x\to \infty\\y\to\infty}\frac{x}{y}$ fails to exist. $\endgroup$
    – Mark Viola
    Nov 6, 2020 at 5:00

1 Answer 1


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).

  • $\begingroup$ I feared that I would have this issue. If you have more time, I have one more question: is there a way to express the integral at infinity as a Riemann sum? $\endgroup$
    – person
    Nov 6, 2020 at 5:15
  • $\begingroup$ For certain functions, the double limit you describe for the improper integral converges to the same number no matter how you take the two indices to infinity. For some it doesn't, and such functions are not integral on $[0,\infty)$. $\endgroup$ Nov 6, 2020 at 5:21

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