as the function is continous at x=π
[tex] \lim_{x \to \ \pi+ } \frac{ \sqrt{2+cos(x)}-1 }{( \pi -x)^2}= \lim_{x \to \ \pi- } \frac{ \sqrt{2+cos(x)}-1 }{( \pi -x)^2}=f( \pi )=k [/tex]
so just find the limit
[tex] \lim_{x \to \ \pi } \frac{ (\sqrt{2+cos(x)}-1)(\sqrt{2+cos(x)}+1) }{( \pi -x)^2(\sqrt{2+cos(x)}+1)} \\ = \lim_{x \to \ \pi } \frac{ (1+cos(x) }{( \pi -x)^2(\sqrt{2+cos(x)}+1)}= \\ \lim_{x \to \ \pi }\frac{ (1+cos(x) }{( \pi -x)^2} \lim_{x \to \ \pi } \frac{1}{(\sqrt{2+cos(x)}+1)} = \\ \lim_{x \to \ \pi }\frac{ (1+cos(x) }{( \pi -x)^2} \frac{1}{2} [/tex]
now we have to apply l-hospital's rule
as it is in form 0/0
[tex]\lim_{x \to \ \pi }\frac{- sin(x) }{-4( \pi -x)} =\lim_{x \to \ \pi } \frac{sin(x)}{4( \pi -x)} [/tex]
again we have to apply l-hospital's rule
as it is in form 0/0
[tex]\lim_{x \to \ \pi } \frac{sin(x)}{4( \pi -x)} =\lim_{x \to \ \pi } \frac{cos(x)}{4(-1)}= \frac{-1}{-4}= \frac{1}{4} [/tex]
