mabey the confusing part is just my font, i need to fix that
2006-09-21 18:17:45 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
lets solve the function first
((3+h)^-1 - (3^-1))/h
=((1/(3+h) - 1/3))/h
= 1/h [(3-3-h)/3(3+h)]
=1/h [-h/3(3+h)]
= -1/3(3+h)
now if h tends to zero, the value of the function will be
= -1/3(3+0)
= -1/9
and that is the answer .. :)
2006-09-21 18:51:46 · answer #1 · answered by mimi 2 · 2⤊ 0⤋
Work on the numerator:
(3+h)^-1 - (3^-1) = 1/(3+h) - 1/3 = (3-3-h) / (3*(3+h)) = -h / (3*(3+h))
The h cancels with the denominator, so you have the limit of:
-1/(3*(3+h))
which is -1/9
2006-09-21 18:50:03 · answer #2 · answered by Andy D. 2 · 0⤊ 0⤋
The limit is -1/9.
2006-09-21 18:24:40 · answer #3 · answered by SonniS 4 · 0⤊ 0⤋
I got it resolved to -1/(9+3h) so as h approaches 0, it evaluates to -1/9.
2006-09-21 18:45:23 · answer #4 · answered by i_sivan 2 · 0⤊ 0⤋
Expression resolves to -h/((3x(3+h))/h) => -1/9 as h=>0
Noel M
2006-09-21 18:30:42 · answer #5 · answered by Noel M 1 · 0⤊ 0⤋
The answer is 1.
2006-09-21 18:24:47 · answer #6 · answered by The Terminator 2 · 0⤊ 0⤋
3x^-7 y^-a million (i imagine you've 3x to the -4 ability and y to the 2d one ability appropriate?once you distinct exponents, you in reality upload them together You get the above, then placed that equation below a million a million/3x^7y^a million
2016-11-23 14:29:11 · answer #7 · answered by dufrene 4 · 0⤊ 0⤋
[(3 + h)^-1 - (3^-1)]/h
= (3^-1 + h^-1 - 3^-1)/h
=(h^-1)/h
=1/h^2
h^2 = 1
h = 1
2006-09-21 18:29:27 · answer #8 · answered by nicel 2 · 0⤊ 0⤋