Bài 3 trang 79

Đề bài

Tìm các giới hạn sau:

a) $\underset{x\to +\infty }{lim}\frac{4x+3}{2x}$;      

b) $\underset{x\to -\infty }{lim}\frac{2}{3x+1}$;

c) $\underset{x\to +\infty }{lim}\frac{\sqrt{x^2+1}}{x+1}$.

Lời giải chi tiết

a) $\underset{x\to +\infty }{lim}\frac{4x+3}{2x}=\underset{x\to +\infty }{lim}\frac{x\left(4+\frac{3}{x}\right)}{2x}=\underset{x\to +\infty }{lim}\frac{4+\frac{3}{x}}{2}=\frac{\underset{x\to +\infty }{lim}4+\underset{x\to +\infty }{lim}\frac{3}{x}}{\underset{x\to +\infty }{lim}2}=\frac{4+0}{2}=2$

b) $\underset{x\to -\infty }{lim}\frac{2}{3x+1}=\underset{x\to -\infty }{lim}\frac{2}{x\left(3+\frac{1}{x}\right)}=\underset{x\to -\infty }{lim}\frac{1}{x}.\underset{x\to -\infty }{lim}\frac{2}{3+\frac{1}{x}}=\underset{x\to -\infty }{lim}\frac{1}{x}.\frac{\underset{x\to -\infty }{lim}2}{\underset{x\to -\infty }{lim}3+\underset{x\to -\infty }{lim}\frac{1}{x}}=0.\frac{2}{3+0}=0$.

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

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