is this the right answer to a LIMIT equation ?

lim x->0 square root of x+3 minus the square root of 3 ALL OVER x

= lim x->0 [sqrt(x+3) -sqrt(3) ]/x = f'(3),

where f(z) =sqrt(x)

f'(z) = 1/2 (x^{-1/2})

so

f'(3) = 1/2(3)^1/2

=sqrt(3) /2sqrt(3)sqrt(3)

=sqrt(3)/6

yes

your answer is correct!!!!

I'm fairly sure the answer is 0....:

lim(x->0) [ sqrt(x+3) - sqrt(3) ] / x

The original equation will lead to 0/0 so we have to use L'Hopital's rule:

H

= [ (1/2)(x+3)^(-1/2) - (1/2)(3)^(-1/2) ] / 1

H

= [ (1/2)(3)^(-1/2) - (1/2)(3)^(-1/2) ] / 1

H

= 0 / 1 = 0

first, rationalize numerator.

then substitute 0 for x.

limit is Sqrt[3]/6. (YES).

lim_x->0 [sqrt(x+3)-sqrt(3)] / x = [x+3-3]/[x(sqrt(x+3)+sqrt(3))] = 1/[sqrt(3)+sqrt(3)] = 1/[2sqrt(3)] = sqrt(3)/6 correct

yes that would be the answer

In short... NO!