Đáp án:
a. $\lim\limits_{x\to7}\dfrac{2-\sqrt{x-3}}{x^2-49}=-\dfrac{1}{56}$
b. $\lim\limits_{x\to0}\dfrac{\sqrt{1+x}-\sqrt{1-x}}{x}=1$
c. $\lim\limits_{x\to-1}\dfrac{\sqrt{2x+3}-\sqrt{x+2}}{3x+3}=\dfrac{1}{6}$
d. $\lim\limits_{x\to7}\dfrac{\sqrt{2x+7}-3}{2-\sqrt{x+3}}=\dfrac{-2(2+\sqrt{10})}{\sqrt{21}+3}$
e) $\lim\limits_{x\to4}\dfrac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=-\dfrac{1}{3}$
Giải thích các bước giải:
a. $\lim\limits_{x\to7}\dfrac{2-\sqrt{x-3}}{x^2-49}=\lim\limits_{x\to7}\dfrac{(2-\sqrt{x-3})(2+\sqrt{x-3})}{(x-7)(x+7)(2+\sqrt{x-3})}$
$=\lim\limits_{x\to7}\dfrac{4-(x-3)}{(x-7)(x+7)(2+\sqrt{x-3})}=\lim\limits_{x\to7}\dfrac{7-x}{(x-7)(x+7)(2+\sqrt{x-3})}$
$=\lim\limits_{x\to7}\dfrac{-1}{(x+7)(2+\sqrt{x-3})}=\dfrac{-1}{(7+7)(2+\sqrt{7-3})}=-\dfrac{1}{56}$
b. $\lim\limits_{x\to0}\dfrac{\sqrt{1+x}-\sqrt{1-x}}{x}=\lim\limits_{x\to0}\dfrac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{x(\sqrt{1+x}+\sqrt{1-x})}$
$=\lim\limits_{x\to0}\dfrac{(1+x)-(1-x)}{x(\sqrt{1+x}+\sqrt{1-x})}=\lim\limits_{x\to0}\dfrac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$
$=\lim\limits_{x\to0}\dfrac{2}{\sqrt{1+x}+\sqrt{1-x}}=\dfrac{2}{1+1}=1$
c. $\lim\limits_{x\to-1}\dfrac{\sqrt{2x+3}-\sqrt{x+2}}{3x+3}=\lim\limits_{x\to-1}\dfrac{(\sqrt{2x+3}-\sqrt{x+2})(\sqrt{2x+3}+\sqrt{x+2})}{3(x+1)(\sqrt{2x+3}+\sqrt{x+2})}$
$=\lim\limits_{x\to-1}\dfrac{(2x+3)-(x+2)}{3(x+1)(\sqrt{2x+3}+\sqrt{x+2})}=\lim\limits_{x\to-1}\dfrac{x+1}{3(x+1)(\sqrt{2x+3}+\sqrt{x+2})}$
$=\lim\limits_{x\to-1}\dfrac{1}{3(\sqrt{2x+3}+\sqrt{x+2})}=\dfrac{1}{3(1+1)}=\dfrac{1}{6}$
d. $\lim\limits_{x\to7}\dfrac{\sqrt{2x+7}-3}{2-\sqrt{x+3}}$
$=\lim\limits_{x\to7}\dfrac{(2x+7-9)(2+\sqrt{x+3})}{(4-x-3)(\sqrt{2x+7}+3)}$
$=\lim\limits_{x\to7}\dfrac{2(x-1)(2+\sqrt{x+3})}{(1-x)(\sqrt{2x+7}+3)}$
$=\lim\limits_{x\to7}\dfrac{-2(2+\sqrt{x+3})}{\sqrt{2x+7}+3}=\dfrac{-2(2+\sqrt{7+3})}{\sqrt{2.7+7}+3}$
$=\dfrac{-2(2+\sqrt{10})}{\sqrt{21}+3}$
e) $\lim\limits_{x\to4}\dfrac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\lim\limits_{x\to4}\dfrac{[9-(5+x)](1+\sqrt{5-x})}{[1-(5-x)](3+\sqrt{5+x})}$
$=\lim\limits_{x\to4}\dfrac{(4-x)(1+\sqrt{5-x})}{(x-4)(3+\sqrt{5+x})}$
$=\dfrac{-(1+\sqrt{5-4})}{3+\sqrt{5+4}}=-\dfrac{1}{3}$
