Giải thích các bước giải:
$\lim_{x\to 1}\dfrac{\sqrt{1+3x}+\sqrt[3]{1-9x}}{(x-1)^2}$
$=\lim_{x\to 1}\dfrac{(\sqrt{1+3x}-(\dfrac{3}{4}x+\dfrac{5}{4}))+((\dfrac{3}{4}x+\dfrac{5}{4})+\sqrt[3]{1-9x})}{(x-1)^2}$
$=\lim_{x\to 1}\dfrac{\dfrac{1+3x-(\dfrac{3}{4}x+\dfrac{5}{4})^2}{\sqrt{1+3x}+(\dfrac{3}{4}x+\dfrac{5}{4})}+\dfrac{1-9x+(\dfrac{3}{4}x+\dfrac{5}{4})^3}{(\dfrac{3}{4}x+\dfrac{5}{4})^2-(\dfrac{3}{4}x+\dfrac{5}{4}).\sqrt[3]{1-9x}+\sqrt[3]{1-9x}^2}}{(x-1)^2}$
$=\lim_{x\to 1}\dfrac{\dfrac{-\dfrac{9}{16}(x-1)^2}{\sqrt{1+3x}+(\dfrac{3}{4}x+\dfrac{5}{4})}+\dfrac{27(x-1)^2(x+7)}{(\dfrac{3}{4}x+\dfrac{5}{4})^2-(\dfrac{3}{4}x+\dfrac{5}{4}).\sqrt[3]{1-9x}+\sqrt[3]{1-9x}^2}}{(x-1)^2}$
$=\lim_{x\to 1}\dfrac{-\dfrac{9}{16}}{\sqrt{1+3x}+(\dfrac{3}{4}x+\dfrac{5}{4})}+\dfrac{27(x+7)}{(\dfrac{3}{4}x+\dfrac{5}{4})^2-(\dfrac{3}{4}x+\dfrac{5}{4}).\sqrt[3]{1-9x}+\sqrt[3]{1-9x}^2}$
$=\dfrac{1143}{64}$
