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Ta có
$\ln L=\lim_{x \to 0} \dfrac{1}{3x+\sin x}\ln\bigg(\frac{x^2+1}{2x^2-x+1}\bigg)$
$\ln L=\lim_{x \to 0} \dfrac{1}{3x+\sin x}\ln\bigg(1+\frac{-x^2+x}{2x^2-x+1}\bigg)$
$\ln L=\lim_{x \to 0} \dfrac{1}{3x+\sin x}.\frac{-x^2+x}{2x^2-x+1}.\frac{2x^2-x+1}{-x^2+x}.\ln\bigg(1+\frac{-x^2+x}{2x^2-x+1}\bigg)$
Đặt $t=\frac{-x^2+x}{2x^2-x+1}$ thì $\lim_{x \to 0}t=0$ tức là khi $x \to 0$ thì $t \to 0$. Ta có
$\ln L = \lim_{x \to 0} \left ( \dfrac{1}{3x+\sin x}.\frac{-x^2+x}{2x^2-x+1} \right ).\lim_{t \to 0}\frac{\ln(t+1)}{t}$
$\ln L = \lim_{x \to 0} \left ( \dfrac{1}{3+\frac{\sin x}{x}}.\frac{-x+1}{2x^2-x+1} \right ).1$
$\ln L = \lim_{x \to 0} \left ( \dfrac{1}{3+1}.\frac{-0+1}{2.0^2-0+1} \right )=\dfrac{1}{4}$
Vậy $L=\sqrt[4]{e}.$
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