# A question from limit

1. Sep 29, 2005

### necessary

:!!)

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

2. Sep 29, 2005

### mathman

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

3. Sep 30, 2005

### necessary

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

(I dont like trigonometric questions in limit)
thanks:)

4. Sep 30, 2005

### 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,

5. Sep 30, 2005

### necessary

thanks,yu are a good man