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Giới hạn cơ bản: $\mathop {\lim }\limits_{x \to 0}\frac{\ln(1+x)}{x}=1$
Ta có:
$\mathop {\lim }\limits_{x \to +\infty}x^2\ln\frac{x^2-1}{x^2+1}$
$=\mathop {\lim }\limits_{x \to +\infty}x^2\ln(1-\frac{2}{x^2+1})$
$=\mathop {\lim }\limits_{x \to +\infty}\frac{\ln(1-\frac{2}{x^2+1})}{-\frac{2}{x^2+1}}.\frac{-2x^2}{x^2+1}=-2$
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