$4cosx.cos2x.cos3x=2.cos3x(cos3x+cosx)=2cos^23x.2cosx.cos3x$$=1+cos6x+cos4x+cos2x$
đề bài $=\mathop {\lim \limits_{x \to 0} \frac{4-1-cos6x-cos4x-cos2x}{1-cosx}}$
$=\mathop {\lim \limits_{x \to 0} \frac{1-cos6x+1-cos4x+1-cos2x}{1-cosx}}$
$=\mathop {\lim \limits_{x \to 0} \frac{\frac{1-cos6x+1-cos4x+1-cos2x}{x^2}}{\frac{1-cosx}{x^2}}} (1)$
Ta có $\mathop {\lim }\limits_{x \to 0} \frac{1-cosax}{x^2}=\frac{a^2}{2}$ (tự cm đc nha bạn, dễ lắm )
Thế vào ta có
$(1)= \frac{\frac{6^2}2 +\frac{4^2}2+\frac{2^2}2}{\frac{1}2}=56$