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I want to evaluate the limit $$\lim_{n\to\infty}n\int_{1-\frac{1}{n}}^{1}f(x)dx.$$

I have seen that the solution to this is $f(1)$, but I don't see how. My idea was to use L'Hospistal's rule, but that didn't work.

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  • $\begingroup$ I think that Riemann sums would help. $\endgroup$ – Polygon Sep 16 '16 at 18:54
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    $\begingroup$ Is $f$ continuous ? $\endgroup$ – Rene Schipperus Sep 16 '16 at 18:54
  • $\begingroup$ Are there any other assumptions on $f$? $\endgroup$ – user305860 Sep 16 '16 at 18:55
  • $\begingroup$ @ReneSchipperus Indeed, $f\in C^{0}[0,1]$. $\endgroup$ – Jason Born Sep 16 '16 at 18:57
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There is a mean value theorem for integrals. $$\frac{1}{b-a}\int_a^bfdx=f(c)$$ for some $c\in [a,b]$.

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Just use the Fundamental theorem of integral calculus: if we set $\;F(x)=\displaystyle\int_1^xf(x)\,\mathrm d\mkern1mu x$, and $f$ is continuous, then $F$ is differentiable and $\;F'(x)=f(x)$.

In the present case, we consider the function $G(x)=\displaystyle\int_{1-x}^1f(x)\,\mathrm d\mkern1mu x=-F(1-x)$. By composition, $$G'(x)= -F'(1-x)\cdot(-1)=f(1-x).$$ In particular, since $G(0)=0$, $$G'(0)=f(1)=\lim_{n\to\infty}\frac{G\bigl(\frac1n\bigr)}{\frac1n}=\lim_{n\to\infty}n\int_{1-\tfrac1n}^1f(x)\,\mathrm d\mkern1mu x. $$

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\begin{align*} \lim_{n\rightarrow \infty} n\int_{1-\frac{1}{n}}^1 f(x) dx &=\lim_{n\rightarrow \infty}\frac{\int_0^1f(x) dx - \int_0^{1-\frac{1}{n}}f(x) dx}{\frac{1}{n}}\\ &=\frac{d\left(\int_0^t f(x)dx\right)}{dt}\Big|_{t=1}\\ &=f(1). \end{align*}

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  • $\begingroup$ This is a very nice solution indeed! I just felt like the other one was simpler, but thanks. $\endgroup$ – Jason Born Sep 16 '16 at 19:09
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you can use L'HOpital just express tje integral as

$$ \frac{F(1)-F(1-1/n)}{1/n} $$

here $ F(X)$ is the integral of $ f(x) $ if i use l'hopital on n i get

$$ -f(1-1/n)(1/n^2 )(-n^2)= f(1) $$ after cancelling terms $n \sim \infty$

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