$\lim_{x\rightarrow 0}\dfrac{1-cosx.cos3x.cos4x}{x^{2}}$
$=\lim_{x\rightarrow 0}\dfrac{1-cosx}{x^{2}}+\lim_{x\rightarrow 0}\dfrac{cosx(1-cos3x)}{x^{2}}+\lim_{x\rightarrow 0}\dfrac{cosx.cos3x(1-cos4x)}{x^{2}}$
$=\lim_{x\rightarrow 0}\dfrac{sin^{2}\frac{x}{2}}{\left ( \frac{x}{2} \right )^{2}.2}+\lim_{x\rightarrow 0}\dfrac{cosx.sin^{2}\frac{3x}{2}}{\left ( \frac{3x}{2} \right )^{2}.\frac{2}{9}}+\lim_{x\rightarrow 0}\dfrac{cosx.cos3x.sin^{2}2x}{(2x)^{2}.\frac{1}{8}}$
$=\lim_{\frac{x}{2}\rightarrow 0}\dfrac{1}{2}.\left ( \dfrac{sin\frac{x}{2}}{\frac{x}{2}} \right )^{2}+\lim_{\frac{3x}{2}\rightarrow 0}\dfrac{9}{2}cosx.\left ( \dfrac{sin\frac{3x}{2}}{\frac{3x}{2}} \right )^{2}+\lim_{2x\rightarrow 0}8cosx.cos3x.\left ( \dfrac{sin2x}{2x} \right )^{2}$
$=\dfrac{1}{2}+\dfrac{9}{2}+8=13$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
