Bài 5.21 trang 123

Đề bài

Cho hàm số $f\left(x\right)=\sqrt{x+1}-\sqrt{x+2}$. Mệnh đề đúng là

A. $\underset{x\to +\infty }{lim}f\left(x\right)=-\infty$                     

B. $\underset{x\to +\infty }{lim}f\left(x\right)=0$             

C. $\underset{x\to +\infty }{lim}f\left(x\right)=-1$    

D. $\underset{x\to +\infty }{lim}f\left(x\right)=-\frac{1}{2}$

Lời giải chi tiết

Ta có:

$\underset{x\to +\infty }{lim}\left(\sqrt{x+1}-\sqrt{x+2}\right)$

$=\underset{x\to +\infty }{lim}\frac{\left(\sqrt{x+1}-\sqrt{x+2}\right)\left(\sqrt{x+1}+\sqrt{x+2}\right)}{\sqrt{x+1}+\sqrt{x+2}}$

$=\underset{x\to +\infty }{lim}\frac{x+1-x-2}{\sqrt{x+1}+\sqrt{x+2}}$

$=\underset{x\to +\infty }{lim}\frac{-1}{\sqrt{x+1}+\sqrt{x+2}}$

$=\underset{x\to +\infty }{lim}\frac{-1}{x\sqrt{1+\frac{1}{x}}+x\sqrt{1+\frac{2}{x}}}$

$=\underset{x\to +\infty }{lim}\frac{\frac{-1}{x}}{\sqrt{1+\frac{1}{x}}+\sqrt{1+\frac{2}{x}}}=\frac{0}{2}=0$.

Đáp án: B.