Giải thích các bước giải:
Bài 23:
$\lim _{x\to 0}\dfrac{\sqrt{x+1}+\sqrt{x+4}-3}{x}$
$=\lim _{x\to 0}\dfrac{\sqrt{x+1}-1+\sqrt{x+4}-2}{x}$
$=\lim _{x\to 0}\dfrac{\dfrac{x+1-1}{\sqrt{x+1}+1}+\dfrac{x+4-4}{\sqrt{x+4}+2}}{x}$
$=\lim _{x\to 0}\dfrac{\dfrac{x}{\sqrt{x+1}+1}+\dfrac{x}{\sqrt{x+4}+2}}{x}$
$=\lim _{x\to 0} \dfrac{1}{\sqrt{x+1}+1}+\dfrac{1}{\sqrt{x+4}+2}$
$= \dfrac{1}{\sqrt{0+1}+1}+\dfrac{1}{\sqrt{0+4}+2}$
$= \dfrac{3}{4}$
