Bài 5.28 trang 124

Đề bài

Tính các giới hạn sau:

a) $\underset{x\to 7}{lim}\frac{\sqrt{x+2}-3}{x-7}$;                   

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

c) $\underset{x\to 1}{lim}\frac{2-x}{{\left(1-x\right)}^2}$;             

d) $\underset{x\to -\infty }{lim}\frac{x+2}{\sqrt{4x^2+1}}$.

Lời giải chi tiết

a) $\underset{x\to 7}{lim}\frac{\sqrt{x+2}-3}{x-7}=\underset{x\to 7}{lim}\frac{1}{\sqrt{x+2}+3}=\frac{1}{6}$.

b) $\underset{x\to 1}{lim}\frac{x^3-1}{x^2-1}=\underset{x\to 1}{lim}\frac{x^2+x+1}{x+1}=\frac{3}{2}$.

c) $\underset{x\to 1}{lim}\frac{2-x}{{\left(1-x\right)}^2}$

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

$\underset{x\to 1}{lim}\left(2-x\right)=1$.

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

d) $\underset{x\to -\infty }{lim}\frac{x+2}{\sqrt{4x^2+1}}=\underset{x\to -\infty }{lim}\frac{1+\frac{2}{x}}{\sqrt{4+\frac{1}{x^2}}}=-\frac{1}{2}$.