lim-0-sin-cos-sin-

Question Number 121813 by bemath last updated on 12/Nov/20

\lim_{θ \to 0} \frac{\sin θ - θ \cos θ}{\sin θ - θ} = ?

Answered by bobhans last updated on 12/Nov/20

\lim_{θ \to 0} \frac{\sin θ - θ \cos θ}{\sin θ - θ} =

\lim_{θ \to 0} \frac{θ - \frac{θ^{3}}{6} - θ (1 - \frac{θ^{2}}{2})}{θ - \frac{θ^{3}}{6} - θ} =

\lim_{θ \to 0} \frac{(\frac{2 θ^{3}}{6})}{- (\frac{θ^{3}}{6})} = - 2 .

Answered by liberty last updated on 12/Nov/20

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

\lim_{θ \to 0} \frac{θ \sin θ}{- 2 \sin^{2} (θ / 2)} = \frac{1}{- 2 (\frac{1}{4})} = - 2 . ▴

Answered by Dwaipayan Shikari last updated on 12/Nov/20

\lim_{x \to 0} \frac{x - \frac{x^{3}}{3!} - x + \frac{x^{3}}{2!}}{x - \frac{x^{3}}{6} - x} = \lim_{x \to 0} \frac{\frac{x^{3}}{3}}{- \frac{x^{3}}{6}} = - 2