Limits

Given that $1+cos(\pi) = (\pi-x) = 0$ , the limit will be in Indeterminate Form of $\frac{0}{0}$ . We will try to eliminate this form by multiplying with $\frac{1-cosx}{1-cosx}$ , then the function will be $\frac{1-cos^2(x)}{(\pi-x)^2(1-cosx)} = \frac{sin^2(x)}{(1-cosx)(\pi-x)^2}$

Provided that $\lim_{x\to0}\frac{sin(x)}{x} = 1$ and $sin(\pi-x) = sin(x)$ (proof will not shown here), we will rewrite the function as $\frac{sin^{2}(\pi-x)}{(\pi-x)^2(1-cosx)}$

Therefore

$lim_{x\to\pi}\frac{sin^{2}(\pi-x)}{(\pi-x)^2(1-cosx)} = 1^2\times\frac{1}{1-(-1)} = \frac{1}{2}$

L'Hospital's Rule says that after evaluating a limit and getting an indeterminate form, you can take the derivatives of the numerator and denominator of the function and then take the limit of f'(x)/g'(x) and still arrive at the right answer. For this problem, you need to take the derivatives of the top and bottom twice.

First, lim x-> pi ((1+cosx)/(pi-x)^2)) => 0/0,

lim x-> pi (-sinx)/(-2(pi-x)) = (1/2)(sin(x))/(pi-x)) which yields 0/0 again.

Once more, (1/2) lim x--> pi (cos(x))/((-1)) = (1/2)

Just evaluate the limit at the numerator's and denominator's second derivatives .