\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}-1}{x}\) bằng
\(\dfrac{1}{2}\).\(\dfrac{1}{3}\).\(-\dfrac{1}{2}\).\(\dfrac{1}{4}\).Hướng dẫn giải:\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}-1}{x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{x+1}-1\right)\left(\sqrt{x+1}+1\right)}{x}\) \(=\lim\limits_{x\rightarrow0}\dfrac{x}{x\left(\sqrt{x+1}+1\right)}=\lim\limits_{x\rightarrow0}\dfrac{1}{\sqrt{x+1}+1}=\dfrac{1}{2}\)
