Question

Solve ${lim}_{\theta \to 0}\left(\frac{1-\cos 4\theta }{1-\cos 6\theta }\right)$

Answer 1

We have,

${lim}_{\theta \to 0}\left(\frac{1-\cos 4\theta }{1-\cos 6\theta }\right)$

We know that

$\cos 2x=1-2{\sin }^{2}x$

Therefore,

${lim}_{\theta \to 0}\left(\frac{1-1+2{\sin }^{2}2\theta }{1-1+2{\sin }^{2}3\theta }\right)$

${lim}_{\theta \to 0}\left(\frac{{\sin }^{2}2\theta }{{\sin }^{2}3\theta }\right)$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

${lim}_{\theta \to 0}\left(\frac{2\sin 2\theta \cos 2\theta \left(2\right)}{2\sin 3\theta \cos 3\theta \left(3\right)}\right)$

${lim}_{\theta \to 0}\left(\frac{2\sin 4\theta }{3\sin 6\theta }\right)$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

${lim}_{\theta \to 0}\left(\frac{2\cos 4\theta \left(4\right)}{3\cos 6\theta \left(6\right)}\right)$

$={lim}_{\theta \to 0}\left(\frac{8\cos 4\theta }{18\cos 6\theta }\right)$

$=\frac{8\cos 0}{18\cos 0}$

$=\frac{8}{18}$

$=\frac{4}{9}$

Hence, this is the answer.