Giải thích các bước giải:
Ta có:
$\lim_{x\to0}\dfrac{1-\sqrt{ax+1}}{\sin(bx)}$
$=\lim_{x\to0}\dfrac{\dfrac{1-(ax+1)}{1+\sqrt{ax+1}}}{\sin(bx)}$
$=\lim_{x\to0}\dfrac{\dfrac{-ax}{1+\sqrt{ax+1}}}{\sin(bx)}$
$=\lim_{x\to0}\dfrac{\dfrac{-a}{1+\sqrt{ax+1}}}{\dfrac{\sin(bx)}{x}}$
$=\lim_{x\to0}\dfrac{\dfrac{-a}{b(1+\sqrt{ax+1})}}{\dfrac{\sin(bx)}{bx}}$
$=\dfrac{\dfrac{-a}{b(1+\sqrt{a\cdot 0+1})}}{1}$
$=-\dfrac{a}{2b}$