Câu b. ta có $\dfrac{\sqrt{x+\sqrt{x^2+1}}-1}{x} =\dfrac{\sqrt{x^2+1} +(x-1)}{x. [ \sqrt{x+\sqrt{x^2+1}}+1]}$
$=\dfrac{2x}{x. [ \sqrt{x+\sqrt{x^2+1}}+1].[\sqrt{x^2+1}-(x-1)]} =\dfrac{2}{ [ \sqrt{x+\sqrt{x^2+1}}+1].[\sqrt{x^2+1}-(x-1)]}$
Vậy $\lim \limits_{x\to 0} \dfrac{2}{ [ \sqrt{x+\sqrt{x^2+1}}+1].[\sqrt{x^2+1}-(x-1)]}=\dfrac{1}{2}$