|
|
Nếu ý em ở đây $\lim$ là $\mathop {\lim }\limits_{x \to 0}$
$\mathop {\lim }\limits \frac{1-\sqrt{1+4x^2}}{1-\cos 2x}=\mathop {\lim }\limits \frac{1-\sqrt{1+4x^2}}{x^2}.\mathop {\lim }\limits \frac{x^2}{1-\cos 2x}$
$=\mathop {\lim }\limits \frac{-4}{1+\sqrt{1+4x^2}}.\mathop {\lim }\limits \frac{x^2}{2\sin^2x}=(-2).\dfrac{1}{2}=-1$
|