Tính \(\lim\limits_{x\rightarrow0}\frac{1-\cos3x.\cos5x.\cos7x}{\sin^27x}\) .
\(\frac{83}{49}\) \(\frac{83}{98}\) \(\frac{81}{49}\) \(\frac{81}{98}\) Hướng dẫn giải:Có \(1-\cos3x\cos5x\cos7x=1-\cos3x+\cos3x\left(1-\cos5x\right)+\cos3x\cos5x\left(1-\cos7x\right)\)
\(=\frac{\sin^23x}{1+\cos3x}+\cos3x.\frac{\sin^25x}{1+\cos5x}+\cos3x\cos5x.\frac{\sin^27x}{1+\cos7x}\)
nên \(\frac{1-\cos3x\cos5x\cos7x}{\sin^27x}=\left(\frac{\sin3x}{\sin7x}\right)^2.\frac{1}{1+\cos3x}+\left(\frac{\sin5x}{\sin7x}\right)^2\frac{\cos3x}{1+\cos5x}+\frac{\cos3x\cos5x}{1+\cos7x}\)
Chú ý rằng \(\lim\limits_{x\rightarrow0}\frac{\sin nx}{\sin mx}=\lim\limits_{x\rightarrow0}\left(\frac{\sin nx}{nx}:\frac{\sin mx}{mx}.\frac{n}{m}\right)=1.1.\frac{n}{m}=\frac{n}{m}\) và \(\lim\limits_{x\rightarrow0}\cos kx=1\) suy ra giới hạn cần tính bằng \(\left(\frac{3}{7}\right)^2.\frac{1}{2}+\left(\frac{5}{7}\right)^2.\frac{1}{2}+\frac{1}{2}=\frac{1}{2}\left(\frac{9}{49}+\frac{25}{49}+1\right)=\frac{83}{98}\)