Question

giới hạn \(\lim\limits_{x\to +∞} f(x)=\dfrac{\sqrt{x^2+2}-2}{x-2}\) 

Answer 1

 

Answer 2

\(\lim\limits_{x\rightarrow+\infty}f\left(x\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+2}-2}{x-2}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2\left(1+\dfrac{2}{x^2}\right)}-2}{x\left(1-\dfrac{2}{x}\right)}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x\cdot\sqrt{1+\dfrac{2}{x^2}}-2}{x\left(1-\dfrac{2}{x}\right)}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{2}{x^2}}-\dfrac{2}{x}}{1-\dfrac{2}{x}}=\dfrac{\sqrt{1+0}-0}{1-0}=\dfrac{1}{1}=1\)