Question

$\underset{\theta \to 0}{lim}\frac{1-\cos 4\theta }{1-\cos 6\theta }$

Answer 1

$\underset{\theta \to 0}{Lt}\frac{1-\cos 4\theta }{1-\cos 6\theta }$

Applying L-Hospitals rule

$\underset{\theta \to 0}{Lt}\frac{+4\sin 4\theta }{+6\sin 6\theta }$

Applying L-Hospitals rule

$=\underset{\theta \to 0}{Lt}\frac{16\cos 4\theta }{36\cos 6\theta }=\frac{16}{36}\frac{\cos \theta }{\cos \theta }$

$=\frac{4}{9}$.