help please: Lim x->pi (sinx)/(1-x/pi)^2?

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help please: Lim x->pi (sinx)/(1-x/pi)^2?

2007-11-23 11:22:35 · 4 answers · asked by yani 2 in Science & Mathematics ➔ Mathematics

4 answers

use l'hopital's rule
lim = lim(x -> pi) cos(x)/(-2/pi(1-x/pi)) = infinity

2007-11-23 11:37:04 · answer #1 · answered by holdm 7 · 0⤊ 0⤋

use LÂ´HÃ´pital
= lim cos (x) / 2(1-x/pi)(-1/pi)= infinity ( the sign depends if x ==> pi from the right or the left)

2007-11-23 19:34:04 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋

L'Hopital's Rule (differentiate the top and bottom)
Lim x->pi cos(x) / 2(1-x/pi)(-1/pi)
lim x->pi pi cos(x) /(-2+x/pi)
cos(pi)=-1
=lim x->pi -pi/-1
=pi

2007-11-23 19:42:15 · answer #3 · answered by cidyah 7 · 0⤊ 0⤋

It doesnt look as though this converges. As x->pi the formula goes to infinity!

2007-11-23 19:36:43 · answer #4 · answered by perplexed* 3 · 0⤊ 0⤋