Find this limit using L'Hopital's rule

The limit is from x -> 1

...of (cos(((pi)(x))/2)) / x-1

so the numerator is "cosine of ([pi times x] divided by 2)

And the denominator is "x minus 1" .

I need to use L'Hopital's rule to calculate it. The answer is supposed to be zero, but I got -pi/2 . Can someone show me their work?

I got the derivative of the numerator and denominator, which lead me to sin((pi)(x)/2)*(-pi/2) / 1
Since sin of pi/2 is 1, that simplified to 1*(-pi/2)/1 = -pi/2

This is what I got, but according to the book it's wrong.

Can anyone show how their work to get zero, or tell me what I did wrong?

10 points to Best Answer.

Thanks

lim x--> 1 cos(πx/2) / (x - 1)

lim x--> 1 [-sin(πx/2) * (2π/4)] / (1)

lim x--> 1 [-sin(π/2) * π/2] = -1 *π/2

= -π/2

You have the correct answer. I am an applied math major and I am willing to be my house is it -π/2. That is how confident I am =p.

EDIT: Check this out, it can not be wrong:

Copy and past this and put one on the website link I put for you below:
"(cos((pi*x)/2))/(x-1)"

As x goes to: 1

http://www.wolframalpha.com/widgets/gall…

Your book is wrong.

And lim x--> -1 (cos((pi*x)/2))/(x-1) = 0, try it yourself. It is not +π/2.

lim{x->1} (cos(((pi)(x))/2)) / (x-1), 0/0 type
= lim{x->1} (-sin(((pi)(x))/2))(pi/2)
= -pi/2
You got the right answer.

are you sure of the limit point ?...as x ---> -1 you get the [ 0 ]...as typed you are valid