lim-x-pi-sin-x-2-1-cos-sin-x-1-

Question Number 160178 by cortano last updated on 25/Nov/21

\lim_{x \to π} \frac{\sin \frac{x}{2} - 1}{\cos (\sin x) - 1} = ?

Commented by blackmamba last updated on 25/Nov/21

\lim_{x \to π} \frac{\sin \frac{x}{2} - 1}{\cos (\sin x) - 1}

= \lim_{x \to π} \frac{\sin \frac{x}{2} - 1}{- 2 \sin^{2} (\frac{\sin x}{2})}

l e t h = \sin \frac{x}{2}; h \to 1

= \lim_{h \to 1} \frac{h - 1}{- 2 \sin^{2} (h \sqrt{1 - h^{2}})}

= - \frac{1}{2} \lim_{h \to 1} \frac{h - 1}{h^{2} (1 - h) (1 + h)}

= - \frac{1}{2} \times - \frac{1}{2} = \frac{1}{4} .

Answered by FongXD last updated on 25/Nov/21

L = \lim_{x \to π} \frac{\cos (\frac{π - x}{2}) - 1}{\cos (\sin (π - x)) - 1}

let t = π - x, if x \to π, \Rightarrow t \to 0

L = \lim_{t \to 0} \frac{\cos \frac{t}{2} - 1}{\cos (sint) - 1}

L = \lim_{t \to 0} \frac{\frac{1 - \cos \frac{t}{2}}{{(\frac{t}{2})}^{2}} \times \frac{1}{4}}{\frac{1 - \cos (sint)}{\sin^{2} t} \times \frac{\sin^{2} t}{t^{2}}}

L = \frac{\frac{1}{2} \times \frac{1}{4}}{\frac{1}{2} \times 1^{2}} = \frac{1}{4}

Facebook Tweet Pin