lim .... (x^2 - 4)/(x^3 - 8)
x=>2
2006-09-03 08:45:28 · 5 answers · asked by chris 2 in Science & Mathematics ➔ Mathematics
Use L'Hospital rule (as a 0/0 condition is present)
Taking the derivative wrt x of the numerator and denominator separately,
Numerator: d/dx(x^2-4) = 2x
Denominator: d/dx(x^3-8) = 3x
Taking this ratio as x goes to 2 = 2x/3x = 2/3
Please work on your own so you can understand the principles involved.
Good luck.
2006-09-03 08:51:21 · answer #1 · answered by alrivera_1 4 · 2⤊ 0⤋
The previous answer is incorrect, the differentiation was done wrong. See corrected version below:
Use L'Hospital rule (as a 0/0 condition is present)
Taking the derivative wrt x of the numerator and denominator separately,
Numerator: d/dx(x^2-4) = 2x
Denominator: d/dx(x^3-8) = 3x^2
Taking this ratio as x goes to 2 = 2x/3x^2 = 2/6 = 1/3
Q.E. D.
2006-09-03 08:56:34 · answer #2 · answered by Answers1 6 · 1⤊ 0⤋
lim((x^2 -4)/(x^3 -8))
x=>2
factor the numberator and the denominator
x^-4 = (x-2)(x+2)
x^-8 = (x-2)(x^2 + 2x + 4)
therefore lim[((x-2)(x+2)) / ((x-2)(x^2 + 2x + 4))]
cancel out common terms
so lim((x+2)/ (x^2 +2x + 4))
x=>2
subs x = 2
4 / 4 + 4 + 4
or 1 / 3
2006-09-03 09:11:34 · answer #3 · answered by Anonymous · 0⤊ 0⤋
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2016-11-24 20:03:32 · answer #4 · answered by ? 4 · 0⤊ 0⤋
factorize
(x-2)(x+2) / (x-2)(x^2 + ...), cross away the (x-2) factor fill in x = 2 and voila the answer...
2006-09-03 08:57:41 · answer #5 · answered by gjmb1960 7 · 0⤊ 0⤋