Tìm giới hạn _{x

* Hướng dẫn giải

\(\begin{array}{l} \text { Ta có: } \sqrt{x^{2}+x+1}-2 \sqrt{x^{2}-x}+x=\dfrac{\left(\sqrt{x^{2}+x+1}+x\right)^{2}-4\left(x^{2}-x\right)}{\sqrt{x^{2}+x+1}+2 \sqrt{x^{2}-x}+x} \\ =\dfrac{2 x \sqrt{x^{2}+x+1}+1+5 x-2 x^{2}}{\sqrt{x^{2}+x+1}+2 \sqrt{x^{2}-x}+x} \\ =\dfrac{2 x\left(\sqrt{x^{2}+x+1}-x\right)}{\sqrt{x^{2}+x+1}+2 \sqrt{x^{2}-x}+x}+\frac{1+5 x}{\sqrt{x^{2}+x+1}+2 \sqrt{x^{2}-x}+x} \\ =\dfrac{2 x(x+1)}{\left(\sqrt{x^{2}+x+1}+2 \sqrt{x^{2}-x}+x\right)\left(\sqrt{x^{2}+x+1}+x\right)} \quad+\dfrac{1+5 x}{\sqrt{x^{2}+x+1}+2 \sqrt{x^{2}-x}+x} \end{array}\)

\(\begin{array}{l} \text { Do đó: } A=\lim \limits_{x \rightarrow+\infty} \frac{2+\dfrac{2}{x}}{\left(\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^{2}}}+2 \sqrt{1-\dfrac{1}{x}}+1\right)\left(\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^{2}}}+1\right)}+ \lim \limits_{x \rightarrow+\infty} \dfrac{\frac{1}{x}+5}{\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^{2}}}+2 \sqrt{1-\dfrac{1}{x}}+1}\\ =\dfrac{1}{4}+\dfrac{5}{4}=\dfrac{3}{2} \end{array}\)