Bài 5.12 trang 118

Đề bài

Tính các giới hạn sau:

a) $\underset{x\to +\infty }{lim}\frac{1-2x}{\sqrt{x^2+1}}$;

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

Lời giải chi tiết

Vì $x\to +\infty$ nên $x>0$, suy ra $\left|x\right|=x$.

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

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

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

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

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

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

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

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

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

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

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

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

$=\frac{1+0}{\sqrt{1+0+0}+1}=\frac{1}{2}$.