$\lim_{x \to 3} $ $\frac{\sqrt[]{x-3}-x+3 }{x-3}$ Lam ho minh voi ah 21/10/2021 Bởi Hadley $\lim_{x \to 3} $ $\frac{\sqrt[]{x-3}-x+3 }{x-3}$ Lam ho minh voi ah
$I=\lim\limits_{x\to 3}\dfrac{ \sqrt{x-3}-(x-3) }{x-3}$ $=\lim\limits_{x\to 3}\dfrac{ (x-3)-(x-3)^2}{(x-3)(\sqrt{x-3}+(x-3) )}$ $=\lim\limits_{x\to 3}\dfrac{1-(x-3) }{\sqrt{x-3}+(x-3)}$ $=+\infty$ (do $x-3>0$) Bình luận
$I=\lim\limits_{x\to 3}\dfrac{ \sqrt{x-3}-(x-3) }{x-3}$
$=\lim\limits_{x\to 3}\dfrac{ (x-3)-(x-3)^2}{(x-3)(\sqrt{x-3}+(x-3) )}$
$=\lim\limits_{x\to 3}\dfrac{1-(x-3) }{\sqrt{x-3}+(x-3)}$
$=+\infty$ (do $x-3>0$)