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$L=\lim_{x\to0}\left ( \frac{1}{(1+x)x}-\dfrac{\ln(1+x)}{x^2} \right )=\lim_{x\to0}\frac{x-(1+x)\ln(1+x)}{(1+x)x^2}$
Áp dụng liên tiếp quy tắc L'Hospital ta được
$L=\lim_{x\to0}\frac{\left[ {x-(1+x)\ln(1+x)} \right]'}{\left[ {(1+x)x^2} \right]'}=-\lim_{x\to0}\frac{\ln(1+x)}{3x^2+2x}$
$L=\lim_{x\to0}\frac{\left[ {\ln(1+x)} \right]'}{\left[ {3x^2+2x} \right]'}=-\lim_{x\to0}\frac{1}{(x+1)(6x+2)}=-\frac{1}{2}$.
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