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what is the limit of [(x)^(1/3) - (x)^(1/5)] / [(x)^(1/3) + (x)^(1/5)] as x approaches infinity?

can you show me the steps?!
thanks

2006-12-09 17:29:30 · 5 answers · asked by *hi* 4 in Science & Mathematics Mathematics

5 answers

[(x)^(1/3) - (x)^(1/5)] / [(x)^(1/3) + (x)^(1/5)] =
[(x)^(1/5)/(x)^(1/3) - 1] / [(x)^(1/5)/(x)^(1/3) + 1] = F
Since (x)^(1/5)/(x)^(1/3) tends to 0 if x grows infinitely
the limit of F is -1

Th

2006-12-09 17:42:50 · answer #1 · answered by Thermo 6 · 0 0

[x^(1/3) - x^(1/5)] / [x^(1/3) + x^(1/5)]

Take out the factor x^(1/5) from
both numerator and denominator.

Then you get :

[x^(5/3) - 1] / [x^(5/3) + 1]

Now divide each term by x^(5/3). Then you get :

[1 - 1/x^(5/3)] / [1 + 1/x^(5/3)]

Now, as x tends to infinity,
x^(5/3) tends to infinity,
and 1/x^(5/3) tends to zero.

Therefore, the last expression becomes :
(1 - 0) / (1 + 0)
= 1/ 1
= 1
So the limit is 1.

2006-12-10 02:05:24 · answer #2 · answered by falzoon 7 · 0 0

Divide the top and bottom by x^(1/3)

The limit is 1

2006-12-10 01:36:30 · answer #3 · answered by z_o_r_r_o 6 · 0 0

you have to multiply the limit by (x^(1/3)+X^(1/5))/(x^(1/3)+x^(1/5)) and you will get lim (x-x)/(x^(1/3)+x^(1/5))^2, which equals zero

2006-12-10 01:40:28 · answer #4 · answered by Jared S 2 · 0 0

~I could, but then I would feel guilty about helping you cheat with your homework and you wouldn't learn anything. The former I can do something about. It appears the latter is a lost cause.

2006-12-10 01:32:35 · answer #5 · answered by Anonymous · 0 0

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