Question

Lim theta-->0 of sin6theta cubed over 7theta

Answer 1

`\displaystyle\lim_(\theta\to0)(\sin^36\theta)/(7\theta)`

`=\displaystyle\lim_(\theta\to0)((6\theta)^3\sin^36\theta)/(7\theta(6\theta)^3)`

`=\displaystyle\lim_(\theta\to0)((6\theta)^3)/(7\theta)*\lim_(\theta\to0)(\sin^36\theta)/((6\theta)^3)`

`=\displaystyle\frac{6^3}7\lim_(\theta\to0)\theta^2*\left(\lim_(\theta\to0)(\sin6\theta)/(6\theta)\right)^3`

`=\displaystyle\frac{6^3}7*0*1^3`

`=0`