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$f$ is a continuous function from $(0,\infty) \to R$ with $f(x) \to a$ as $x \to \infty$. Can I write:

$$\lim_{X\rightarrow \infty}\lim_{n\rightarrow \infty}\sum_{i=1}^{n}f(X\cdot i/n) \cdot (1/n)= a$$

So basically, I am taking the limit on $X \to \infty$ inside the sum as it is a finite sum for a fixed $n$. That should be allowed, I believe.

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You can't generally switch the order of two limits, and what you have written there amounts to taking $n$ out to infinity while $X$ is held fixed first, and then taking $X$ to infinity, though I think some analysis will show it does come out to be $a$ this way. If you try the limits the other way it's rather trivial that you will get $a$ as the limit. – anon Feb 16 '12 at 21:20

$$\lim_{X\to\infty}\lim_{n\to\infty} \sum_{i=1}^n f(X\cdot i/n)\cdot 1/n=\lim_{X\to\infty} \frac{1}{X}\int_0^X f(u)du=\lim_{X\to\infty} f(X)=a \tag{1}$$

$$\lim_{n\to\infty}\lim_{X\to\infty} \sum_{i=1}^n f(X\cdot i/n)\cdot 1/n=\lim_{n\to\infty} \sum_{i=1}^n a/n=\lim_{n\to\infty} a=a \tag{2}$$

The double limit in $(2)$ is straightforward: $\lim\limits_{X\to\infty}f(X\cdot i/n)=a$ for each summand. However you cannot evaluate the version in $(2)$ and automatically say it equals the version in $(1)$, because in general you cannot interchange the order of two limits, e.g. $\lim\limits_x \lim\limits_y\, (x/y)$. However, since $f\to a$ in the limit, we have that $f\in (a-\epsilon,a+\epsilon)$ for any $\epsilon>0$ for all $X>N$ for some real $N$, which means

$$\left|\frac{1}{X-N}\int_N^X f(u)du-a\right|<\epsilon $$

ultimately implying

$$\frac{1}{X}\int_0^X f(u)du=a+\mathcal{O}(X^{-1})\to a. $$

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Why your second equality in (1) holds? – leo Feb 17 '12 at 4:57
@leo: That's explained in the very last sentence. Do you want me to add more detail? – anon Feb 17 '12 at 5:30

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