Finding the Limit of a Complex Expression

daniel_i_l
Gold Member
Messages
864
Reaction score
0

Homework Statement



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

Homework Equations





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.
 
Physics news on Phys.org
Check your l'Hopital's rule again. I get 2/pi, not pi/2.
 
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top