$\lim_{x->0} \frac{\sqrt[4]{x+1}-1}{\sqrt[3]{x+1}-1}$
$\lim_{x->0} \frac{\sqrt[4]{x+1}-1}{\sqrt[3]{x+1}-1}$
$=\frac{\sqrt[3]{(x+1)^{2}}+\sqrt[3]{x+1}+1}{\sqrt[4]{(x+1)^{3}}+\sqrt{x+1}+\sqrt[4]{x+1}+1}=\frac{3}{4}$
Đặt $t=\sqrt[4]{x+1}$
$L=\lim_{t\to 1}\frac{t-1}{t^{4/3}-1}=\lim_{t\to 1}\frac{\left (t-1 \right )(t^{8/3}+t^{4/3}+1)}{t^4-1}=\lim_{t\to 1}\frac{t^{8/3}+t^{4/3}+1}{1 + t + t^2 + t^3}=3/4$
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