Ta có bổ đề :$$Lim_{x \rightarrow 0} \frac{\ln(x+cos2x)-x}{sin(x)^2}$$
$ \lim_{x \rightarrow 0} \frac{ f(x)}{g(x)}= $$ \lim_{x \rightarrow 0} \frac{ f'(x)}{g'(x)}$ nếu $f(0) = g(0) = 0 $
Thật vậy :
$ \lim_{x \rightarrow 0} \frac{ f(x)}{g(x)}$$ = \lim_{x \rightarrow 0} \frac{ \frac{f(x) -f(0)}{x-0}}{\frac{g(x)-g(0)}{x-0}}$$ =\lim_{x \rightarrow 0} \frac{ f'(x)}{g'(x)}$
nên :
$L = \lim_{x \rightarrow 0}\frac{\frac{1-2sin2x}{x+cos2x}-1}{sin2x} = \lim_{x \rightarrow 0}\frac{(\frac{1-2sin2x}{x+cos2x}-1)'}{(sin2x)'} =\lim_{x \rightarrow 0}\frac{5-4 x cos(2 x)+4 sin(2 x)}{(x+cos(2 x))^2 (2 cos(2 x))}=\frac{-5}{2} $
- ThaiSon2929 yêu thích