Đáp án:
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos 3x - \cos 6x} }}{{{x^2}}} = + \infty \]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos 3x - \cos 6x} }}{{{x^2}}}\\
\mathop {\lim }\limits_{x \to 0} \sqrt {\cos 3x - \cos 6x} = \sqrt {\cos \left( {3.0} \right) - \cos \left( {6.0} \right)} = 0\\
\Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {1 - \sqrt {\cos 3x - \cos 6x} } \right) = 1 - 0 = 1\\
\left. \begin{array}{l}
\mathop {\lim }\limits_{x \to 0} {x^2} = {0^2} = 0\\
{x^2} \ge 0,\,\,\,\,\forall x
\end{array} \right\} \Rightarrow \mathop {\lim }\limits_{x \to 0} {x^2} = {0^ + }\\
\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos 3x - \cos 6x} }}{{{x^2}}} = + \infty
\end{array}\)
