lim x->0 (x^2/ 1-cos(x)). Solution Recall: sin(0) = 0 cos(0) = 1 If you \"plug i

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Recall: sin(0) = 0 cos(0) = 1 If you "plug in" x=0 you get the form "0 / 0." This means you can apply L'Hopital's rule. So you can take the derivative of top and bottom without changing the limits: [lim x->0] (1-cos x) / x^2 = [lim x->0] - (-sin x) / 2x = [lim x->0] sin x / 2x But if you "plug in" x=0 you get the form "0 / 0" again. So apply the rule once more: [lim x->0] (1-cos x) / x^2 = [lim x->0] sin x / 2x = [lim x->0] cos x / 2 Now you can really plug in x=0 to get 1/2 as the limit. Remember, since we have a real legitimate number here, we can stop -- in fact, we aren't even allowed to keep going with L'Hopital's rule even if we wanted to, since it is not 0/0 anymore. You can conclude: [lim x->0] (1-cos x) / x^2 = [lim x->0] cos x / 2 = cos(0) / 2 = 1/2 so answr is 2

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