Question-155639

Question Number 155639 by cortano last updated on 03/Oct/21

Commented by yeti123 last updated on 03/Oct/21

\lim_{θ \to 0} \frac{\sin 2 θ}{1 - \cos θ} = \lim_{θ \to 0} \frac{\sin 2 θ}{{2 \sin}^{2} (θ / 2)}

= \lim_{θ \to 0} (\frac{\sin 2 θ}{2 θ} \times \frac{{(θ / 2)}^{2}}{{2 \sin}^{2} (θ / 2)} \times \frac{2 θ}{{(θ / 2)}^{2}})

= 1 \times \frac{1}{2} \times \lim_{θ \to 0} \frac{8}{θ}

= \infty

Commented by cortano last updated on 03/Oct/21

the limit {doesn}^{'} t exist

Commented by cortano last updated on 03/Oct/21

\lim_{x \to 0^{+}} \frac{1}{x} = \infty

\lim_{x \to 0^{-}} \frac{1}{x} = - \infty

Commented by yeti123 last updated on 03/Oct/21

= 1 \times \frac{1}{2} \times \lim_{θ \to 0} \frac{8}{θ}

= \lim_{θ \to 0} \frac{4}{θ}

= two - sided limit doesnt exist

Answered by Ar Brandon last updated on 03/Oct/21

= \lim_{ϑ \to 0} \frac{4 ϑ (1 - \frac{ϑ}{2})}{ϑ^{2}} = \lim_{x \to 0} (\frac{4}{ϑ} - 2) \to infinity