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Đề bài

Tìm các giới hạn sau:

a) $\underset{x\to -2}{lim}\left(x^2-7x+4\right)$;

b) $\underset{x\to 3}{lim}\frac{x-3}{x^2-9}$;                              

c) $\underset{x\to 1}{lim}\frac{3-\sqrt{x+8}}{x-1}$.

Lời giải chi tiết

a) $\underset{x\to -2}{lim}\left(x^2-7x+4\right)$

$=\underset{x\to -2}{lim}\left(x^2\right)-\underset{x\to -2}{lim}\left(7x\right)+\underset{x\to -2}{lim}4$

$=\underset{x\to -2}{lim}\left(x^2\right)-7\underset{x\to -2}{lim}x+\underset{x\to -2}{lim}4$

$={\left(-2\right)}^2-7.\left(-2\right)+4=22$.

b) $\underset{x\to 3}{lim}\frac{x-3}{x^2-9}=\underset{x\to 3}{lim}\frac{x-3}{\left(x-3\right)\left(x+3\right)}=\underset{x\to 3}{lim}\frac{1}{x+3}$

$=\frac{\underset{x\to 3}{lim}1}{\underset{x\to 3}{lim}x+\underset{x\to 3}{lim}3}=\frac{1}{3+3}=\frac{1}{6}$.

c) $\underset{x\to 1}{lim}\frac{3-\sqrt{x+8}}{x-1}=\underset{x\to 1}{lim}\frac{\left(3-\sqrt{x+8}\right)\left(3+\sqrt{x+8}\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}$

$=\underset{x\to 1}{lim}\frac{3^2-\left(x+8\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\underset{x\to 1}{lim}\frac{1-x}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}$

$=\underset{x\to 1}{lim}\frac{-\left(x-1\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\underset{x\to 1}{lim}\frac{-1}{3+\sqrt{x+8}}$

$=\frac{\underset{x\to 1}{lim}\left(-1\right)}{\underset{x\to 1}{lim}3+\underset{x\to 1}{lim}\sqrt{x+8}}=\frac{-1}{3+\sqrt{1+8}}=-\frac{1}{6}$.