Question

Limit tends to 0 1-cos4x/1-cos6x

Answer 1

`\displaystyle\lim_(x\to0)(1-\cos4x)/(1-\cos6x)=\lim_(x\to0)(1-\cos4x)/(1-\cos6x)*(1+\cos4x)/(1+\cos4x)*(1+\cos6x)/(1+\cos6x)`

`=\displaystyle\lim_(x\to0)(1-\cos^24x)/(1-\cos^26x)*(1+\cos6x)/(1+\cos4x)`

`=\displaystyle\lim_(x\to0)(\sin^24x)/(\sin^26x)*(1+\cos6x)/(1+\cos4x)`

`=\displaystyle\lim_(x\to0)(\sin^24x)/(\sin^26x)*((4x)^2)/((4x)^2)*((6x)^2)/((6x)^2)*(1+\cos6x)/(1+\cos4x)`

`=\displaystyle\lim_(x\to0)\left((\sin4x)/(4x)\right)^2\left((6x)/(\sin6x)\right)^2\left((4x)/(6x)\right)^2(1+\cos6x)/(1+\cos4x)`

`=\displaystyle\left(\lim_(x\to0)(\sin4x)/(4x)\right)^2\left(\lim_(x\to0)(6x)/(\sin6x)\right)^2\left(\lim_(x\to0)(4x)/(6x)\right)^2\lim_(x\to0)(1+\cos6x)/(1+\cos4x)`

Recall that
`\displaystyle\lim_(x\to0)(\sin ax)/(ax)=\lim_(x\to0)(ax)/(\sin ax)=1`. You then have


`\displaystyle\left(\lim_(x\to0)(4x)/(6x)\right)^2\lim_(x\to0)(1+\cos6x)/(1+\cos4x)`

`=\displaystyle\left(\lim_(x\to0)\frac23\right)^2\lim_(x\to0)(1+\cos6x)/(1+\cos4x)`

`=\displaystyle\frac49*(1+1)/(1+1)`

`=\displaystyle\frac49`