lim-x-pi-2-cosx-1-sinx-

Question Number 128439 by Study last updated on 07/Jan/21

l i \underset{x \to \frac{π}{2}}{m} \frac{c o s x}{1 - s i n x} = ? ? ?

Answered by benjo_mathlover last updated on 07/Jan/21

let x = \frac{π}{2} + z \land z \to 0

\lim_{z \to 0} \frac{\cos (\frac{π}{2} + z)}{1 - \sin (z + \frac{π}{2})} = \lim_{z \to 0} \frac{- \sin z}{1 - \cos z}

= \lim_{z \to 0} \frac{- \sin z}{2 \sin^{2} (z / 2)} = - \infty

Answered by liberty last updated on 08/Jan/21

\lim_{x \to π / 2} \frac{\cos^{2} (x / 2) - \sin^{2} (x / 2)}{\sin^{2} (x / 2) - 2 \sin (x / 2) \cos (x / 2) + \cos^{2} (x / 2)}

= \lim_{x \to π / 2} \frac{(\cos \frac{x}{2} - \sin \frac{x}{2}) (\cos \frac{x}{2} + \sin \frac{x}{2})}{{(\cos \frac{x}{2} - \sin \frac{x}{2})}^{2}}

= \lim_{x \to π / 2} \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{∣ \cos \frac{x}{2} - \sin \frac{x}{2} ∣} = \frac{\sqrt{2}}{0} = \infty

or \lim_{x \to π / 2} \frac{\cos (\frac{x}{2}) + \sin (\frac{x}{2})}{∣ \cos \frac{x}{2} - \sin \frac{x}{2} ∣} = - \infty

Answered by MJS_new last updated on 08/Jan/21

limit {doesn}^{'} t exist because

\lim_{x \to \frac{π}{2}^{-}} \frac{\cos x}{1 - \sin x} = + \infty but \lim_{x \to \frac{π}{2}^{+}} \frac{\cos x}{1 - \sin x} = - \infty

Commented by liberty last updated on 08/Jan/21

yes \dots . agree

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