Respuesta :

LammettHash
[tex]\displaystyle\lim_{x\to0}\frac{1-\cos4x}{1-\cos6x}=\lim_{x\to0}\frac{1-\cos4x}{1-\cos6x}\times\frac{1+\cos4x}{1+\cos4x}\times\frac{1+\cos6x}{1+\cos6x}[/tex]
[tex]=\displaystyle\lim_{x\to0}\frac{1-\cos^24x}{1-\cos^26x}\times\frac{1+\cos6x}{1+\cos4x}[/tex]
[tex]=\displaystyle\lim_{x\to0}\frac{\sin^24x}{\sin^26x}\times\frac{1+\cos6x}{1+\cos4x}[/tex]
[tex]=\displaystyle\lim_{x\to0}\frac{\sin^24x}{\sin^26x}\times\frac{(4x)^2}{(4x)^2}\times\frac{(6x)^2}{(6x)^2}\times\frac{1+\cos6x}{1+\cos4x}[/tex]
[tex]=\displaystyle\lim_{x\to0}\left(\frac{\sin4x}{4x}\right)^2\left(\frac{6x}{\sin6x}\right)^2\left(\frac{4x}{6x}\right)^2\frac{1+\cos6x}{1+\cos4x}[/tex]
[tex]=\displaystyle\left(\lim_{x\to0}\frac{\sin4x}{4x}\right)^2\left(\lim_{x\to0}\frac{6x}{\sin6x}\right)^2\left(\lim_{x\to0}\frac{4x}{6x}\right)^2\lim_{x\to0}\frac{1+\cos6x}{1+\cos4x}[/tex]

Recall that [tex]\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=\lim_{x\to0}\frac{ax}{\sin ax}=1[/tex]. You then have

[tex]\displaystyle\left(\lim_{x\to0}\frac{4x}{6x}\right)^2\lim_{x\to0}\frac{1+\cos6x}{1+\cos4x}[/tex]
[tex]=\displaystyle\left(\lim_{x\to0}\frac23\right)^2\lim_{x\to0}\frac{1+\cos6x}{1+\cos4x}[/tex]
[tex]=\displaystyle\frac49\times\frac{1+1}{1+1}[/tex]
[tex]=\displaystyle\frac49[/tex]

Otras preguntas