$\lim_{n \to 0} ( \dfrac{1}{x}$ + $\dfrac{1}{x^2}$ ) = $\lim_{n \to 0} \dfrac{x^2+x}{x^3}$ = $\lim_{n \to 0} \dfrac{\dfrac{1}{x}+\dfrac{1}{x^2}}{1}$ = $∞$ Bình luận
$\lim\limits_{x\to 0}(\dfrac{1}{x}+\dfrac{1}{x^2})$ $=\lim\limits_{x\to 0}\dfrac{x^2+x}{x^3}$ $=\lim\limits_{x\to 0}\dfrac{1+\dfrac{1}{x}}{x}$ $=\dfrac{1+0}{0}=+\infty$ Bình luận
$\lim_{n \to 0} ( \dfrac{1}{x}$ + $\dfrac{1}{x^2}$ )
= $\lim_{n \to 0} \dfrac{x^2+x}{x^3}$
= $\lim_{n \to 0} \dfrac{\dfrac{1}{x}+\dfrac{1}{x^2}}{1}$
= $∞$
$\lim\limits_{x\to 0}(\dfrac{1}{x}+\dfrac{1}{x^2})$
$=\lim\limits_{x\to 0}\dfrac{x^2+x}{x^3}$
$=\lim\limits_{x\to 0}\dfrac{1+\dfrac{1}{x}}{x}$
$=\dfrac{1+0}{0}=+\infty$