limit X->0 (sqrt(3+x)-sqrt(3))/x
Clueless.... Answer is sqrt(3)/6 no idea how to get there
How do i do it without the L'Hôpital's rule. We have not gotten there yet in class. Thanks
If you know the rule of L'Hopital, it becomes easy. The limits of both numerator and denominator are both 0, so we can use L'Hopital:
limit x-->0 (sqrt(3+x) - sqrt(3))/x = limit x-->0 (1/2 1/sqrt(3+x)) / 1 = 1/2 1/sqrt(3) = 1/2 sqrt(3)/3 = sqrt(3)/6
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If you know the rule of L'Hopital, it becomes easy. The limits of both numerator and denominator are both 0, so we can use L'Hopital:
limit x-->0 (sqrt(3+x) - sqrt(3))/x = limit x-->0 (1/2 1/sqrt(3+x)) / 1 = 1/2 1/sqrt(3) = 1/2 sqrt(3)/3 = sqrt(3)/6