toan lop 11

Question

tinh cac gioi han sau

$\mathop {\lim }\limits_{x \to 0}\frac{(x^{2}+2014)\sqrt{1+2x}-2014}{x(4x+5)}$

Thịnh Hải Yến · Answer 1 · 2/23/2014 8:45:53 PM

$=\mathop {\lim }\limits_{x \to 0}\frac{x^{2}\sqrt{1+2x}}{x(4x+5)}+\mathop {\lim }\limits_{x \to 0}\frac{2014(\sqrt{1+2x}-1)}{x(4x+5)}$

$=\mathop {\lim }\limits_{x \to 0}\frac{x\sqrt{1+2x}}{4x+5}+\mathop {\lim }\limits_{x \to 0}\frac{2014(1+2x-1)}{x(4x+5)(\sqrt{1+2x}+1)}$

$=0 + \mathop {\lim }\limits_{x \to 0}\frac{4028x}{x(4x+5)(\sqrt{1+2x}+1)}$

$=\mathop {\lim }\limits_{x \to 0}\frac{4028}{(4x+5)(\sqrt{1+2x}+1)}=\frac{4028}{5.2}=402,5$

1 Đáp án