# A tricky limit

1. Aug 24, 2007

### daniel_i_l

1. The problem statement, all variables and given/known data

Find the limit:
$$lim_{x \rightarrow 1} (2-x)^{tan \frac{\pi x}{2}}$$

2. Relevant equations

3. The attempt at a solution

I first tried to find the limit of ln of the function inorder to turn the power into a multiplication and got:
$$lim_{x \rightarrow 1} \frac{ ln(2-x) sin \frac{\pi x}{2}}{cos \frac{\pi x}{2}}$$
Then I used L'hopitals rule and got:
$$lim_{x \rightarrow 1} ln( (2-x)^{tan \frac{\pi x}{2}} ) = \pi / 2$$
That means that $$lim_{x \rightarrow 1} (2-x)^{tan \frac{\pi x}{2}} = e^{\pi / 2}$$
Is that right? I tried putting in values to my calc and it looks like the answer should be 1?
What did I do wrong?
Thanks.

2. Aug 24, 2007

### Dick

Check your l'Hopital's rule again. I get 2/pi, not pi/2.

3. Aug 24, 2007

### Gib Z

I also get $2/\pi$, and when I put in x=0.9999 on my calculator, the limit and $e^{2/\pi}$ agree to reasonable accuracy, around 1.89, not 1.