Lời giải:
Áp dụng quy tắc $L'Hospital$,ta có:
$lim_{x->0}\frac{\sqrt{x+1}-1}{\sqrt[3]{x+1}-1}$
$=lim_{x->0}\frac{(x+1)^{\frac{1}{2}}-1}{(x+1)^{\frac{1}{3}}-1}$
$=lim_{x->0}\frac{((x+1)^{\frac{1}{2}}-1)'}{((x+1)^{\frac{1}{3}}-1)'}$
$=lim_{x->0}\frac{\frac{1}{2}.(x+1)^{\frac{-1}{2}}}{\frac{1}{3}.(x+1)^{\frac{-2}{3}}}$
$=\frac{\frac{1}{2}.(0+1)^{\frac{-1}{2}}}{\frac{1}{3}.(0+1)^{\frac{-2}{3}}}$
$=\frac{\frac{1}{2}}{\frac{1}{3}}$
$=\frac{3}{2}$
