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The value of lim(x->pi) (sqrt(2+cosx)-1)...

The value of `lim_(x->pi) (sqrt(2+cosx)-1)/(x-pi)^2`

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`lim(x->pi) (sqrt(2+cosx)-1)/(x-pi)^2`
This is a `(0/0)` form
So, apply L- Hospital Rule
`=lim(x->pi) (d/(dx)(sqrt(2+cosx)-1))/(d/(dx)(x-pi)^2)`
`=lim(x->pi) (-sin x)/(4(x-pi)(sqrt(2+cos x)))`
`=lim(x->pi) (-cos x)/(4(sqrt(2+cosx).(1)+(x-pi)(-sin x)/(2 sqrt(2+cosx))))`
`=1/4`
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