Tìm giới hạn _{x x}-2

* Hướng dẫn giải

\(\begin{aligned} \text { Ta có: } & \sqrt{x^{2}+2 x}-2 \sqrt{x^{2}+x}+x=\frac{2 x^{2}+2 x+2 x \sqrt{x^{2}+2 x}-4 x^{2}-4 x}{\sqrt{x^{2}+2 x}+2 \sqrt{x^{2}+x}+x} \\ &=2 x \frac{\sqrt{x^{2}+2 x}-x-1}{\sqrt{x^{2}+2 x}+2 \sqrt{x^{2}+x}+x} \\ &=\frac{-2 x}{\left(\sqrt{x^{2}+2 x}+2 \sqrt{x^{2}+x}+x\right)\left(\sqrt{x^{2}+2 x}+x+1\right)} \end{aligned}\)

\(\begin{array}{l} \text { Nên } B=\lim\limits _{x \rightarrow+\infty} \dfrac{-2 x^{2}}{\left(\sqrt{x^{2}+2 x}+2 \sqrt{x^{2}+x}+x\right)\left(\sqrt{x^{2}+2 x}+x+1\right)} \\ =\lim\limits _{x \rightarrow+\infty} \frac{-2}{\left(\sqrt{1+\dfrac{2}{x}}+2 \sqrt{1+\frac{1}{x}}+1\right)\left(\sqrt{1+\dfrac{2}{x}}+1+\dfrac{1}{x}\right)}=-\dfrac{1}{4} \end{array}\)