evaluate limit x tends to zero (cosx-1/x^2)
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limit (cosx-1)/x^2
x__0
while putting the value of X we get 0/0 form so using L hospital rule
( diffrenciating function f(x) and g(x) where f(x)=cosx-1 and g(x) = x^2
lim(_sin x)/2x
x___0
again 0/0 form so again diffrenciating using L hospital rule
lim(-cos x/2)
x__0
so ans is -1/2
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