Giải thích các bước giải:
Ta có :
$\lim_{x\to0}\dfrac{\sqrt[m]{1+\alpha\:x}-1}{\alpha\:x}=\dfrac{1}{m}$
$\to \sqrt[m]{1 +\alpha x}=1 +\dfrac{\alpha x}m$
$\lim_{x\to0}\dfrac{\sqrt{1+\alpha\:x}\sqrt[3]{1+\beta\:x}\sqrt[4]{1+\gamma\:x}-1}{x}$
$=\lim_{x\to0}\dfrac{(1+\dfrac{\alpha x}{2})(1+\dfrac{\beta x}{3})(1+\dfrac{\gamma x}{4})-1}{x}$
$=\lim_{x\to0}\dfrac{\dfrac{αβx^2}{6}+\dfrac{βx}{3}+\dfrac{αx}{2}+\dfrac{xγ}{4}+\dfrac{αx^2γ}{8}+\dfrac{βx^2γ}{12}+\dfrac{αβx^3γ}{24}}{x}$
$=\lim_{x\to0}\dfrac{αβx}{6}+\dfrac{β}{3}+\dfrac{α}{2}+\dfrac{γ}{4}+\dfrac{αxγ}{8}+\dfrac{βxγ}{12}+\dfrac{αβx^2γ}{24}$
$=\dfrac{β}{3}+\dfrac{α}{2}+\dfrac{γ}{4}$
$\to D$