Using L'Hopital's Rule to Evaluate the Limit: x->pi/2: (4x-2pi)/(cos(2pi-x))

[BigNate]

Hello,

Can someone please help me evaluate the limit?

I'm trying to take the limit as x-> pi/2 of the following:

(4x-2pi)/(cos(2pi-x))

The derivative of the numerator is 4.
The derivative I get for the denominator is -sin(2pi - x) - 1
When I plug in my limit, pi/2, I get -sin(3/2pi)-1 = 0

Am I evaluating the derivative of the denominator correctly? The answer to this problem is -4, but at this point, I have an answer of 4/0 so clearly I am not doing this correctly.

Any help and guidance would be much appreciated! Thanks!

 

[ksdhart2]

Well, you've correctly identified that your error lies in the derivative of the denominator. It's close to the right answer, but I'm not quite sure where the minus 1 term at the end came from. If we let \(\displaystyle u=2\pi-x\) and apply the chain rule, we get:

\(\displaystyle \dfrac{d}{dx} (cos(u))=-sin(u) \cdot \dfrac{du}{dx}\)

Try continuing from there and see what you get. It should work out okay, giving you the desired denominator of -1 when you plug in pi/2.

 

[BigNate]

Thanks! I just made a careless mistake on the derivative. I added du/dx instead od multiplying by du/dx.