Giải thích các bước giải:
e.Ta có:
$\lim_{x\to-1}\dfrac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}$
$=\lim_{x\to-1}\dfrac{\dfrac{x+1}{(\sqrt[3]{x})^2-\sqrt[3]{x}+1}}{\dfrac{x^2+3-2^2}{\sqrt{x^2+3}+2}}$
$=\lim_{x\to-1}\dfrac{\dfrac{x+1}{(\sqrt[3]{x})^2-\sqrt[3]{x}+1}}{\dfrac{x^2-1}{\sqrt{x^2+3}+2}}$
$=\lim_{x\to-1}\dfrac{\dfrac{x+1}{(\sqrt[3]{x})^2-\sqrt[3]{x}+1}}{\dfrac{(x-1)(x+1)}{\sqrt{x^2+3}+2}}$
$=\lim_{x\to-1}\dfrac{\dfrac{1}{(\sqrt[3]{x})^2-\sqrt[3]{x}+1}}{\dfrac{x-1}{\sqrt{x^2+3}+2}}$
$=\dfrac{\dfrac{1}{(\sqrt[3]{-1})^2-\sqrt[3]{-1}+1}}{\dfrac{-1-1}{\sqrt{(-1)^2+3}+2}}$
$=-\dfrac23$
f.Ta có:
$\lim_{x\to -2}\dfrac{x^4-16}{x^2+6x+8}$
$=\lim_{x\to -2}\dfrac{(x-2)(x+2)(x^2+2)}{(x+2)(x+4)}$
$=\lim_{x\to -2}\dfrac{(x-2)(x^2+2)}{x+4}$
$=\dfrac{(-2-2)((-2)^2+2)}{-2+4}$
$=-12$
