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How to find the following limit without evaluating the integral? $\lim_{k\rightarrow \infty }\left(k\int_0^1 (x-1) x^k (\log x)^{-1} \, dx\right)$ , $k>-1$

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What does $\log^{-1}$ mean? – Daniel Littlewood Dec 9 '12 at 17:21
Might it be that $\left(\log x\right)^{-1}$ was intended? – Michael Hardy Dec 9 '12 at 17:30
it means $1/log(x)$ – naanwa Dec 9 '12 at 17:30

Let $f_k(x)=(x-1)x^k(\log(x))^{-1}$. Note then that $f_1\in L^1[0,1]$, $f_k$ is a positive sequence, and $f_{k+1}(x)\leqslant f_k(x)$. So, you can apply the Monotone Convergence Theorem to get that the limit is zero.

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Sorry,I forgot to include a factor of $k$ inside the integral .I fixed it .The limit should be 1. – naanwa Dec 9 '12 at 17:27

It can be shown that $$x < \frac{x-1}{\log x} < 1 \qquad \textrm{for} \qquad 0 < x < 1.$$ (This is related to the first inequality here.) In addition $$\lim_{x\to 0} (x-1)/\log x = 0 \qquad \textrm{and} \qquad \lim_{x\to 1} (x-1)/\log x = 1.$$ Therefore, for $k > 0$ $$\int_0^1 dx\, k x^{k+1} < \int_0^1 dx\, k\frac{(x-1)x^k}{\log x} < \int_0^1 dx\, k x^{k}$$ and so $$\frac{k}{k+2} < \int_0^1 k\frac{(x-1)x^k}{\log x} \, dx < \frac{k}{k+1}.$$ By the squeeze theorem we have $$\lim_{k\to\infty} \int_0^1 k\frac{(x-1)x^k}{\log x} = 1.$$

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