lim x->0 square root of x+3 minus the square root of 3 ALL OVER x
is the right answer square-root3 /6
2006-12-02 11:14:16 · 6 answers · asked by PrimeTyme Berri 2 in Science & Mathematics ➔ Mathematics
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!!!!
2006-12-02 11:21:46 · answer #1 · answered by Anonymous · 1⤊ 0⤋
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
2006-12-02 11:38:30 · answer #2 · answered by Mark 2 · 0⤊ 1⤋
first, rationalize numerator.
then substitute 0 for x.
limit is Sqrt[3]/6. (YES).
2006-12-02 11:19:43 · answer #3 · answered by mr green 4 · 1⤊ 0⤋
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
2006-12-02 11:20:20 · answer #4 · answered by jacinablackbox 4 · 0⤊ 0⤋
yes that would be the answer
2006-12-02 11:18:08 · answer #5 · answered by raj 7 · 0⤊ 0⤋
In short... NO!
2006-12-02 11:16:04 · answer #6 · answered by Anonymous · 0⤊ 3⤋