Bài 3.7 trang 74

Đề bài

Tính các giới hạn sau:

a, $\underset{x\to 0}{lim}\frac{x^2+3x+5}{x+1}$

b, $\underset{x\to 2}{lim}\frac{x^2+x-6}{x^2-4}$

c, $\underset{x\to -2}{lim}\frac{\sqrt{x+11}-3}{x+2}$

d, $\underset{x\to -\infty }{lim}\frac{3x^2+x+10}{2x^2-1}$

e, $\underset{x\to -\infty }{lim}\frac{5x^3+9}{x^4+1}$

g, $\underset{x\to -\infty }{lim}\frac{\sqrt{x^2+1}}{x}$

Lời giải chi tiết

a, Ta có: $\underset{x\to 0}{lim}(x^2+3x+5)=5$ và $\underset{x\to 0}{lim}(x+1)=1$

Vậy $\underset{x\to 0}{lim}\frac{x^2+3x+5}{x+1}=5$

b, Ta có : $f\left(x\right)=\frac{x^2+x-6}{x^2-4}=\frac{(x+3).(x-2)}{(x-2).(x+2)}=\frac{x+3}{x+2}$

$\underset{x\to 2}{lim}(x+3)=5$ và $\underset{x\to 2}{lim}(x+2)=4$

Vậy $\underset{x\to 2}{lim}\frac{x^2+x-6}{x^2-4}=\frac{5}{4}$.

c, Ta có: $f\left(x\right)=\frac{\sqrt{x+11}-3}{x+2}=\frac{(\sqrt{x+11}-3)(\sqrt{x+11}+3)}{x+2}=\frac{x+11-3^2}{x+2}=1$

$\underset{x\to -2}{lim}1=1$

Vậy $\underset{x\to -2}{lim}\frac{\sqrt{x+11}-3}{x+2}=1$

d, Ta có: $\underset{x\to -\infty }{lim}\frac{3x^2+x+10}{2x^2-1}=\underset{x\to -\infty }{lim}\frac{3+\frac{1}{x}+\frac{10}{x^2}}{2-\frac{1}{x^2}}=\frac{3}{2}$

e, Ta có: $\underset{x\to -\infty }{lim}\frac{5x^3+9}{x^4+1}=\underset{x\to -\infty }{lim}\frac{5+\frac{9}{x^4}}{1+\frac{1}{x^4}}=5$

g, Ta có: $\underset{x\to -\infty }{lim}\frac{\sqrt{x^2+1}}{x}=\underset{x\to -\infty }{lim}\frac{\left|x\right|.\sqrt{1+\frac{1}{x^2}}}{x}=\underset{x\to -\infty }{lim}\frac{-x.\sqrt{1+\frac{1}{x^2}}}{x}=\underset{x\to -\infty }{lim}(-\sqrt{1+\frac{1}{x^2}})=-1$.