Quote from chineserider
For the first problem, use l'hopital's rule and take the derivative of the numerator over the derivative of the denominator. The answer to the second problem should be infinity because a real number over zero goes to infinity.
Not entirely sure of the answers though.
Not entirely sure of the answers though.
I really don't think he had l'hopital, looking at these problems and that he'd like to solve it by graphing.
but I have to admit, it IS a pain without...
1st:
lim_x->0 (1-cos x)/sin x
= lim_x->0 (1/2-1/2 cos x)/ (1/2 sin x) multiplied numerator and denominator with 1/2
= lim_x->0 (sin² (x/2))/(sin (x/2)*cos (x/2)) used sin² x=1/2(1-cos 2x) and sin x * cos x = 1/2 sin 2x
=lim_x->0 sin (x/2) / cos (x/2) = lim_x->0 tan (x/2) = 0
for the second you are right.
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