Find the Limit of (1-x)tan(πx/2) as x Approaches 1 - Get Help Here!

[necessary]

:!)

lim(1-x).tan Pi X/2=?
x->1

please help meee :!)

 

[mathman]

Let y=1-x and use tan=sin/cos and also sinu=cos(pi/2 -u), you will get your answer easily.

 

[necessary]

thanks..
one more guestion that I can't solve is...lim 1-sinX/cosx =?
X->pi/2

(I don't like trigonometric questions in limit)
thanks:)

 

[VietDao29]

\lim_{x \rightarrow \frac{\pi}{2}} \frac{1 - \sin x}{\cos x}
You can try multiplying both numerator and denominator by (1 + sin x), something like:
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{(1 - \sin x)(1 + \sin x)}{\cos x(1 + \sin x)}
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{1 - \sin ^ 2 x}{\cos x(1 + \sin x)}
1 - sin²x = ...
Can you go from here?
----------------------
Or you can try a different way:
1 - \sin x = \sin \left( \frac{\pi}{2} \right) - \sin x = 2 \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)
\lim_{x \rightarrow \frac{\pi}{2}} \frac{1 - \sin x}{\cos x}
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{\sin\left( \frac{\pi}{2} - x \right)}
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{2\sin \left( \frac{\pi}{4} - \frac{x}{2} \right) \cos \left( \frac{\pi}{4} - \frac{x}{2} \right)}
= \lim_{x \rightarrow \frac{\pi}{2}} \frac{ \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{\cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \cos \left( \frac{\pi}{4} - \frac{x}{2} \right)}
Viet Dao,

 

[necessary]

thanks,yu are a good man