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Math limits at infinity?

Evaluate the limit.



2. lim \/x^3-2x^2+1]/[x-1]
x-->+infinity
4. lim [\/1+\/x]/ [\/1+x]
x-->+infinity
6. lim [^3\/x^3+2] - [^3\/x^3-1]
x-->-infinity
7. lim[ x^2-1]/ [\/x^2+1]
x-->-infinity
8. lim [\/x^2-2x+3 - \/x]
x-->+infinity
10. lim [\/x^3-4x^2+1] + [\/x^2-x] +x

2007-07-20 01:29:25 · 5 answers · asked by jan 2 in Science & Mathematics Mathematics

5 answers

2. lim \/x^3-2x^2+1]/[x-1] = lim x \/x - 2/x2 + 1/x^2]/[{x +1] =
lim x/(x +1) * lim \/x - 2/x2 + 1/x^2
As x --> oo, x/( x +1) --> 1 and x - 2/x2 + 1/x^2 --> oo, which implies \/x - 2/x2 + 1/x^2 --> oo. Therefore,
lim \/x^3-2x^2+1]/[x-1] = oo
x-->+infinity

The other exercises are similar

2007-07-20 02:38:30 · answer #1 · answered by Steiner 7 · 0 0

specially, something that oscillates diverges. by using fact this might nicely be a series, and you at the instant are not including any words, it particularly is going to converge to 0 by using fact the denominator is increasing to countless. It being unfavourable or beneficial wont make a difference by using fact it is so close to to 0.

2016-10-09 03:03:18 · answer #2 · answered by Anonymous · 0 0

2.divid by x
=0

2007-07-20 01:37:42 · answer #3 · answered by Anonymous · 0 0

take derivative of numerator and denominator of each fraction

2007-07-20 01:36:01 · answer #4 · answered by iyiogrenci 6 · 0 1

the limit is infinity....


where on earth can you apply calculus???!!!

2007-07-20 01:35:02 · answer #5 · answered by I waltz in coke 2 · 0 1

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