Đáp án:
\[\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 3x}}{{1 - \cos x}} = 9\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 3x}}{{1 - \cos x}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{1 - \left( {4{{\cos }^3}x - 3\cos x} \right)}}{{1 - \cos x}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{ - 4{{\cos }^3}x + 3\cos x + 1}}{{1 - \cos x}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{4{{\cos }^3}x - 3\cos x - 1}}{{\cos x - 1}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {4{{\cos }^3}x - 4{{\cos }^2}x} \right) + \left( {4{{\cos }^2}x - 4\cos x} \right) + \left( {\cos x - 1} \right)}}{{\cos x - 1}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{4{{\cos }^2}x\left( {\cos x - 1} \right) + 4\cos x\left( {\cos x - 1} \right) + \left( {\cos x - 1} \right)}}{{\cos x - 1}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\cos x - 1} \right)\left( {4{{\cos }^2}x + 4\cos x + 1} \right)}}{{\cos x - 1}}\\
= \mathop {\lim }\limits_{x \to 0} \left( {4{{\cos }^2}x + 4\cos x + 1} \right)\\
= 4.{\cos ^2}0 + 4\cos 0 + 1\\
= 9
\end{array}\)
