Finding Lim Without L'Hospital: A Math Challenge

[medwatt]

Hello.
I have been trying to find this limit:

Lim as x --> 1 of (sin((1-x)/2)*tan(Pi*x/2))

Of course I don't want to solve it using L'Hospital. I have tried several ways but ended up in one of these.

lim as y -->0 of (y*tan(pi/2-y*pi))

The answer when using L'Hospital is 1/pi.

How can I go about getting the same answer without using L'Hospital. Thanks.

 

[mfb]

= $\sin\left(\frac{1-x}{2}\right) \frac{\sin(\pi x/2)}{\cos(\pi x/2)}$
sin(πx/2) goes to 1 and does not matter. With z=x/2-1/2 and z->0 this can be rewritten as
$$ \frac{-\sin\left(z\right)}{\cos(\pi z+\frac{\pi}{2})} = \frac{\sin\left(z\right)}{\sin(\pi z)}$$
You can use common upper and lower bounds for sin() to get the limit of that expression.