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

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

  • 0
    Just to nitpick, maybe the limit should be as $x\to a^{-}$ or else the integral is improper. When x>a, the integral is always infinite or undefined.2012-11-28

4 Answers 4

3

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). $

2

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

2

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

1

Here is general proof without using limit theorems:

First remark that

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

hence

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

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)$

  • 0
    That looks nice also, thanks.2012-11-28