lim ((x+2)^3*(1 -2*x)^4)/((2*x+3)^2*(3 -x)^5) 13/11/2021 Bởi Parker lim ((x+2)^3*(1 -2*x)^4)/((2*x+3)^2*(3 -x)^5)
$\lim\limits_{x\to +\infty}\dfrac{(x+2)^3(1-2x)^4}{(2x+3)^2(3-x)^5}$ $=\lim\limits_{x\to +\infty}\dfrac{x^3(1+\dfrac{2}{x})^3. x^4(\dfrac{1}{x}-2)^4}{ x^2(2+\dfrac{3}{x})^2.x^5(\dfrac{3}{x}-1)^5}$ $=\lim\limits_{x\to +\infty}\dfrac{(1+\dfrac{2}{x})^3(\dfrac{1}{x}-2)^4}{(2+\dfrac{3}{x})^2.(\dfrac{3}{x}-1)^5}$ $=\dfrac{(-2)^4}{2^2.(-1)^5}$ $=-4$ Bình luận
$\lim\limits_{x\to +\infty}\dfrac{(x+2)^3(1-2x)^4}{(2x+3)^2(3-x)^5}$
$=\lim\limits_{x\to +\infty}\dfrac{x^3(1+\dfrac{2}{x})^3. x^4(\dfrac{1}{x}-2)^4}{ x^2(2+\dfrac{3}{x})^2.x^5(\dfrac{3}{x}-1)^5}$
$=\lim\limits_{x\to +\infty}\dfrac{(1+\dfrac{2}{x})^3(\dfrac{1}{x}-2)^4}{(2+\dfrac{3}{x})^2.(\dfrac{3}{x}-1)^5}$
$=\dfrac{(-2)^4}{2^2.(-1)^5}$
$=-4$