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Here's a little question I saw in a book recently, which I can see but can't set out my own formal proof and it's annoying me.

Say $\lim_{x\to\infty} g(x) = a$

and $g$ is continuous,

so now prove that

$\lim_{x\to\infty} \frac{1}{x} \int_0^x g(y) \mathrm{d}y = a$ .

I can see that if we define $\int_0^x g(y)\mathrm{d}y = G(x) - G(0)$

then $\frac{1}{x} \int_0^x g(y) \mathrm{d}y = \frac{G(x) - G(0)}{x}$

which clearly looks a lot like the limit of a differential, but I'm not sure how to handle the limit?!

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    Is $g$ continuous?2012-10-10
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    Yes, we can assume $g$ is continuous.2012-10-10
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    Can you include it in the OP? Thanks!2012-10-10

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