$\lim_{x \to0 }\frac{1-cosx.cos2x.cos3x}{1-cosx}$
$1-cosx.cos2x.cos3x=(1-cosx).cos2x.cos3x+(1-cos2x).cos3x+(1-cos3x)$
Nhận thấy: $\frac{1-coskx}{x^{2}}=\frac{2sin^{2}\frac{kx}{2}}{x^{2}}=\frac{k^{^{2}}}{2}.\frac{sin^{2}\frac{kx}{2}}{\frac{k^{2}.x^{2}}{4}}=\frac{k^{^{2}}}{2}$
Ta có: $\lim_{x\rightarrow 0}cosx=\lim_{x\rightarrow 0}cos2x=\lim_{x\rightarrow 0}cos3x=1$
Nên: $\lim_{x\rightarrow o}\frac{1-cosx.cos2x.cos3x}{1-cosx}=\lim_{x\rightarrow 0}\frac{\frac{1-cosx.cos2x.cos3x}{x^{2}}}{\frac{1-cosx}{x^{2}}}=\lim_{x\rightarrow 0}\frac{\frac{(1-cosx).cos2x.cos3x+(1-cos2x).cos3x+(1-cos3x)}{x^{2}}}{\frac{1-cosx}{x^{2}}}=\lim_{x\rightarrow 0}\frac{\frac{1}{2}+2+\frac{9}{2}}{\frac{1}{2}}=14$
- Ispectorgadget, C a c t u s và Hatmua09 thích