Giải thích các bước giải:
$\lim_{x\to 1}\dfrac{\sqrt{1+3x}+\sqrt[3]{1-9x}}{(x-1)^2}$
$=\lim_{x\to 1}\dfrac{(\sqrt{1+3x}-(\dfrac{3}{4}x+\dfrac{5}{4}))+((\dfrac{3}{4}x+\dfrac{5}{4})+\sqrt[3]{1-9x})}{(x-1)^2}$
$=\lim_{x\to 1}\dfrac{\dfrac{1+3x-(\dfrac{3}{4}x+\dfrac{5}{4})^2}{\sqrt{1+3x}+(\dfrac{3}{4}x+\dfrac{5}{4})}+\dfrac{1-9x+(\dfrac{3}{4}x+\dfrac{5}{4})^3}{(\dfrac{3}{4}x+\dfrac{5}{4})^2-(\dfrac{3}{4}x+\dfrac{5}{4}).\sqrt[3]{1-9x}+\sqrt[3]{1-9x}^2}}{(x-1)^2}$
$=\lim_{x\to 1}\dfrac{\dfrac{-\dfrac{9}{16}(x-1)^2}{\sqrt{1+3x}+(\dfrac{3}{4}x+\dfrac{5}{4})}+\dfrac{27(x-1)^2(x+7)}{(\dfrac{3}{4}x+\dfrac{5}{4})^2-(\dfrac{3}{4}x+\dfrac{5}{4}).\sqrt[3]{1-9x}+\sqrt[3]{1-9x}^2}}{(x-1)^2}$
$=\lim_{x\to 1}\dfrac{-\dfrac{9}{16}}{\sqrt{1+3x}+(\dfrac{3}{4}x+\dfrac{5}{4})}+\dfrac{27(x+7)}{(\dfrac{3}{4}x+\dfrac{5}{4})^2-(\dfrac{3}{4}x+\dfrac{5}{4}).\sqrt[3]{1-9x}+\sqrt[3]{1-9x}^2}$
$=\dfrac{1143}{64}$
b.$\lim_{x\to 0}\dfrac{\sqrt[3]{1+3x}-1-x\sqrt{1-x}}{x^3+x^2}$
$=\lim_{x\to 0}\dfrac{\dfrac{1+3x-1}{\sqrt[3]{1+3x}^2+\sqrt[3]{1+3x}+1}-x\sqrt{1-x}}{x^3+x^2}$
$=\lim_{x\to 0}\dfrac{\dfrac{3x}{3}-x\sqrt{1-x}}{x^3+x^2}$
$=\lim_{x\to 0}\dfrac{x-x\sqrt{1-x}}{x^3+x^2}$
$=\lim_{x\to 0}\dfrac{1-\sqrt{1-x}}{x^2+x}$
$=\lim_{x\to 0}\dfrac{\dfrac{1-(1-x)}{1+\sqrt{1-x}}}{x^2+x}$
$=\lim_{x\to 0}\dfrac{\dfrac{x}{1+\sqrt{1-0}}}{x^2+x}$
$=\lim_{x\to 0}\dfrac{\dfrac{x}{2}}{x^2+x}$
$=\lim_{x\to 0}\dfrac{\dfrac{1}{2}}{x+1}$
$=\dfrac{\dfrac{1}{2}}{0+1}$
$=\dfrac 12$
