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How to prove for a continuous function $f$, the following limit holds?

$$\lim_{x\to a}\,(a-x)\int_0^x\frac{f(y)}{(a-y)^2}\,dy=f(a)$$

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Something like ? – limits Nov 28 '12 at 10:51
I would try to edit the header to be a little more informative. "Limit question" is not so useful for anyone hoping to use the site as a reference. Welcome to the site though, I hope you find it useful! – Simon Hayward Nov 28 '12 at 11:07
Shouldn't the integral go from $x$ to $a$? – Berci Nov 28 '12 at 11:15
Why has this question got 4 up-votes? No research effort, no attempt at an answer, title is vague... – Epictetus Nov 28 '12 at 11:16
That's fine. Look's like you've got some useful answers anyway. Why not use these forums as a chance to practice using Latex? It'll come in handy as you progress as a mathematician :) – Simon Hayward Nov 28 '12 at 11:35
up vote 3 down vote accepted

Use l'Hospital: $$ \lim_{x\to a}\,\frac{\int_0^x\frac{f(y)}{(a-y)^2}\,dy}{\frac1{(a-x)}}= \lim_{x\to a}\,\frac{\frac d{dx}\left(\int_0^x\frac{f(y)}{(a-y)^2}\,dy\right)}{\frac d{dx}\left(\frac1{(a-x)}\right)}= \lim_{x\to a}\,\frac{\frac{f(x)}{(a-x)^2}}{\frac1{(a-x)^2}}=\lim_{x\to a}f(x)=f(a). $$ By that, this should then generalise to $$ \lim_{x\to a}\,\frac{(a-x)^n}{n}\int_0^x\frac{f(y)}{(a-y)^{n+1}}\,dy=f(a). $$

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Thank you very much! – limits Nov 28 '12 at 11:33

Hint: Use the Lagrange Mean Value Theorem for the differentiable $F(x):=\displaystyle\int_0^x \frac{f(y)}{(a-y)^2}dy $.

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Rewrite this limit as $$ \lim\limits_{x\to a}\frac{\int\limits_0^x\frac{f(y)}{(a-y)^2}}{\frac{1}{a-x}} $$ and use L'Hopitale rule

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Thank you very much! – limits Nov 28 '12 at 11:33

Here is general proof without using limit theorems:

First remark that

$$(x-a) \cdot \int_0^x \frac{f(a)}{(a-y)^2} = f(a)+ \frac{f(a) \cdot (x-a)}{a}$$


$$\left|(a-x) \cdot \int_0^x \frac{f(y)}{(a-y)^2} \, dy - f(a) \right| \leq \left| (a-x) \cdot \int_0^x \frac{f(y)}{(a-y)^2} \, dy - f(a) + \frac{f(a) \cdot (x-a)}{a} \right| + \underbrace{\left|\frac{f(a) \cdot (x-a)}{a} \right|}_{\to 0 \, (x \to a)}$$

Now we have by the first equation

$$(a-x) \cdot \int_0^x \frac{f(y)}{(a-y)^2} \, dy - f(a) + \frac{f(a) \cdot (x-a)}{a} = (x-a) \cdot \int_0^x \frac{f(y)-f(a)}{(a-y)^2} \, dy $$

Let $\varepsilon>0$. Since $f$ is continuous we find $\delta>0$ such that $|f(y)-f(a)| \leq \varepsilon$ for all $y \in B(a,\delta)$. Thus

$$\left|(x-a) \cdot \int_{(0,x) \cap B(a,\delta)} \frac{f(y)-f(a)}{(a-y)^2}\right| \leq \left(1- \frac{x-a}{a} \right) \cdot \varepsilon \\ \left|(x-a) \cdot \int_{(0,x) \cap B(a,\delta)^c} \frac{f(y)-f(a)}{(a-y)^2}\right| \leq (x-a) \cdot 2 \|f\|_{\infty} \cdot \frac{x}{\delta^2} \leq (x-a) \cdot 2 \|f\|_{\infty} \cdot \frac{x}{\delta_0^2} $$

for some fixed $\delta_0>0$. Hence

$$ \left| (a-x) \cdot \int_0^x \frac{f(y)}{(a-y)^2} \, dy - f(a) + \frac{f(a) \cdot (x-a)}{a} \right| \leq (x-a) \cdot 2 \|f\|_{\infty} \cdot \frac{x}{\delta_0^2} + \left(1- \frac{x-a}{a} \right) \cdot \varepsilon \to 0 \quad (\varepsilon \to 0, x \to a)$$

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That looks nice also, thanks. – limits Nov 28 '12 at 11:44

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