Answers

2014-04-21T12:43:35+05:30
As the function is continous at x=π
 \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
so just find the limit
 \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}
now we have to apply l-hospital's rule
as it is in form 0/0
\lim_{x \to \ \pi }\frac{- sin(x) }{-4( \pi -x)} =\lim_{x \to \ \pi } \frac{sin(x)}{4( \pi -x)}
again  we have to apply l-hospital's rule
as it is in form 0/0
\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}


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2014-04-27T11:20:19+05:30
The answer of this question is 1/2
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