Giải thích các bước giải:
7.Ta có:
$G=\lim_{x\to-1}\dfrac{\sqrt{x^2+3}-2}{\sqrt[3]{x+9}-2}$
$\to G=\lim_{x\to-1}\dfrac{\dfrac{x^2+3-2^2}{\sqrt{x^2+3}+2}}{\dfrac{x+9-2^3}{(\sqrt[3]{x+9})^2+2\sqrt[3]{x+9}+4}}$
$\to G=\lim_{x\to-1}\dfrac{\dfrac{x^2-1}{\sqrt{x^2+3}+2}}{\dfrac{x+1}{(\sqrt[3]{x+9})^2+2\sqrt[3]{x+9}+4}}$
$\to G=\lim_{x\to-1}\dfrac{\dfrac{(x-1)(x+1)}{\sqrt{x^2+3}+2}}{\dfrac{x+1}{(\sqrt[3]{x+9})^2+2\sqrt[3]{x+9}+4}}$
$\to G=\lim_{x\to-1}\dfrac{\dfrac{x-1}{\sqrt{x^2+3}+2}}{\dfrac{1}{(\sqrt[3]{x+9})^2+2\sqrt[3]{x+9}+4}}$
$\to G=\lim_{x\to-1}\dfrac{(x-1)((\sqrt[3]{x+9})^2+2\sqrt[3]{x+9}+4)}{\sqrt{x^2+3}+2}$
$\to G=\dfrac{(-1-1)((\sqrt[3]{-1+9})^2+2\sqrt[3]{-1+9}+4)}{\sqrt{(-1)^2+3}+2}$
$\to G=-6$
8.Ta có:
$\lim_{x\to0}\dfrac{\sqrt{x+9}+\sqrt{x+16}-7}{x^2-3x}$
$=\lim_{x\to0}\dfrac{(\sqrt{x+9}-3)+(\sqrt{x+16}-4)}{x(x-3)}$
$=\lim_{x\to0}\dfrac{\dfrac{x+9-3^2}{\sqrt{x+9}+3}+\dfrac{x+16-4^2}{\sqrt{x+16}+4}}{x(x-3)}$
$=\lim_{x\to0}\dfrac{\dfrac{x}{\sqrt{x+9}+3}+\dfrac{x}{\sqrt{x+16}+4}}{x(x-3)}$
$=\lim_{x\to0}\dfrac{\dfrac{1}{\sqrt{x+9}+3}+\dfrac{1}{\sqrt{x+16}+4}}{x-3}$
$=\dfrac{\dfrac{1}{\sqrt{0+9}+3}+\dfrac{1}{\sqrt{0+16}+4}}{0-3}$
$=-\dfrac7{72}$
