limit of ((4/x-4/3)/(x-3)) as x approaches 3
i got 4/9 but the answers is -4/9?
why is it negative?
2007-01-28 04:27:36 · 4 answers · asked by SeriousTyro 2 in Science & Mathematics ➔ Mathematics
Because if you use the De L' Ho>pital rule, you have to find the derivative of 4/x, which is -4/x^2. (You have to use the rule, because the fraction ((4/x-4/3)/(x-3)) becomes a 0/0 when you replace x with 3).
2007-01-28 04:34:19 · answer #1 · answered by supersonic332003 7 · 0⤊ 0⤋
differentiating
-4/x^2/1
allowingthe limit
=-4/9
2007-01-28 04:32:59 · answer #2 · answered by raj 7 · 0⤊ 0⤋
lt (4/x-4/3)/(x-3)
x->3
= lt (4/x-4/3)/(x-3)
x->3
= lt (12-4x)/3x(x-3)
x->3
= lt 4(3-x)/3x(x-3)
x->3
= lt - 4(x-3)/3x(x-3)
x->3
= lt - 4/3x
x->3
= -4/9
===
2007-01-28 04:58:50 · answer #3 · answered by george t 2 · 0⤊ 0⤋
lim [ (4/x) - (4/3) ] / (x-3)
= lim [ (4/x) (3/3) - (4/3) (x/x) ] / (x-3)
=lim [ (12/3x) -(4x/3x) / (x-3)
=lim [ (12-4x) / 3x ] / (x-3)
=lim [ (4) (3-x) / 3x ] / (x-3)
=lim [ (4) (-1) / 3x
=lim [ (-4)/ 3x ] = -4/9
x->3
2007-01-28 05:24:43 · answer #4 · answered by ignoramus_the_great 7 · 1⤊ 0⤋