Tính các giới hạn sau: a) lim x->2 căn (4x+1) -3 /x-2

Giải SBT Toán 11 Bài 16: Giới hạn của hàm số

Bài 5.12 trang 83 SBT Toán 11 Tập 1: Tính các giới hạn sau:

a) $\underset{x\to 2}{\lim }\frac{\sqrt{4x+1}-3}{x-2}$ ;

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

c) $\underset{x\to 2^{+}}{\lim }\frac{x^2-5x+6}{{\left(x-2\right)}^2}$ ;

d) $\underset{x\to 0^{-}}{\lim }\frac{x^2+x-2}{x}$ .

Lời giải:

a)$\underset{x\to 2}{\lim }\frac{\sqrt{4x+1}-3}{x-2}$$=\underset{x\to 2}{\lim }\frac{4x+1-3^2}{\left(x-2\right)\left(\sqrt{4x+1}+3\right)}$

$=\underset{x\to 2}{\lim }\frac{4\left(x-2\right)}{\left(x-2\right)\left(\sqrt{4x+1}+3\right)}=\underset{x\to 2}{\lim }\frac{4}{\sqrt{4x+1}+3}=\frac{2}{3}$.

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

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

c) $\underset{x\to 2^{+}}{\lim }\frac{x^2-5x+6}{{\left(x-2\right)}^2}$$=\underset{x\to 2^{+}}{\lim }\frac{\left(x-2\right)\left(x-3\right)}{{\left(x-2\right)}^2}=\underset{x\to 2^{+}}{\lim }\frac{x-3}{x-2}$ .

Vì $\underset{x\to 2^{+}}{\lim }\left(x-2\right)=\mathrm{0,}\underset{x\to 2^{+}}{\lim }\left(x-3\right)=2-3=-1<0$ và x – 2 > 0 khi x → 2⁺, nên $\underset{x\to 2^{+}}{\lim }\frac{x-3}{x-2}=-\infty$.

Vậy $\underset{x\to 2^{+}}{\lim }\frac{x^2-5x+6}{{\left(x-2\right)}^2}=-\infty$ .

d)$\underset{x\to 0^{-}}{\lim }\frac{x^2+x-2}{x}$

Vì $\underset{x\to 0^{-}}{\lim }\left(x^2+x-2\right)=0+0-2=-2<0$ , $\underset{x\to 0^{-}}{\lim }x=0$ và x < 0 nên $\underset{x\to 0^{-}}{\lim }\frac{x^2+x-2}{x}=+\infty$ .