Question

lim h->0 = ((u+h)^(1/3) - u^(1/3))/h

Answer 1

Rationalize the numerator.

Answer 2

how?

Answer 3

Multiply by a factor of 1 to preserve the expression. That "1" takes the form of what you already have for your current numerator, except switch the sign between the terms. Make the denominator the same. That will be a factor of "1", and you will get an "h" to factor out.

Answer 4

don't understand :(

Answer 5

Multiply by ((u+h)^(2/3) + u^(2/3))/((u+h)^(2/3) + u^(2/3)). See what you get.

Answer 6

top and bottom?

Answer 7

You have a "/" in the above factor that I gave to you. That should give you your answer. Plus, this factor I gave to you is exactly the same in the numerator and denominator which means that you are multiplying by a factor of one. You are doing this because "1" is the multiplicative identity. For real numbers, 1 x a = a, which gives you your justification for multiplying by "1". The particular value was chosen to rationalize your exponents.

Answer 8

Your initial expression is in the form a/b. The factor I gave to you is in the form m/m. (a/b) x (m/m) = (am) / (bm).

Answer 9

kay, thank you :D im gonna try it out, thank yo so much
if i still cant do it i might bother you again ;)

Answer 10

It's just simple arithmetic at this point, you shouldn't have any trouble. The trick is to find a suitable factor of 1. 2 things were necessary. 1) working with the exponent and 2) changing the sign. You'll see once you do the arithmetic.

Answer 11

how do you multiply ((u+h)^(2/3))(u^(1/3))

Answer 12

don't. you'll find out when you multiply that you have a ((u+h)^(2/3))(u^(1/3))-((u+h)^(2/3))(u^(1/3)) which cancels out. One is positive and one is negative.

Answer 13

omg i give up can't understand this uu' thanks

Answer 14

use the fact that

\[a^3 – b^3 = (a – b)(a^2 + ab + b^2)\]

Answer 15

You can do that, or break up the "intermediate" answer into the sum of 2 limits. That's actually going to be easier. This is a difficult problem, but I'm sitting here with the answer, and this might be one of those cases, where I should just do all the work. We're not supposed to do all the work, but this time might be an exception. To the other helpers: should I just go ahead?

Answer 16

I would suggest showing most of the steps but don't solve the limit.

Answer 17

Ok, I'll try not to give away the store. I think I can do it.

Answer 18

Thanks you so much :) I'll do everything after what you give me

Answer 19

or use the standard limit :
\(\huge\lim_{x->a} \frac{x^n-a^n}{x-a}=na^{n-1}\)
here put x=u+h

Answer 20

I'm supposed to use the lim->0 f(x+h) - f(x) / h

Answer 21

I got the function of f(x) = u ^ (1/3) so i replaced it

Answer 22

Your "intermediate" limit is [h - u^(1/3)(u+h)^(1/3)[(u+h)^(1/3) - u^(1/3)] / (h[(u+h)^(2/3) + u^(2/3)]). This is 2 terms, with the first term having only an "h" in the numerator. That will factor out with the "h" in the denominator immediately. For the second term, you have to again rationalize that 3rd factor of the second term in the numerator. Another "h" will pop out.

Answer 23

This is all done by separating the intermediate answer first into TWO limits.

Answer 24

It does work, but it's a lot of math. The math is not really difficult, just darn tricky. But it works.

Answer 25

I do have to take someone to the doctor's office, and will be back in about 1-2 hours. I'll log on to see if you need any more help, but you should get it by just foloowing this second rationalizing, or using perhaps one of the other helper's methods. I didn't have time to check their methods, but they're really good and should be able to explain their method or help you work through mine. Mine is actually very fundamental, but a "grind it out" method.

Answer 26

\[a^3 – b^3 = (a – b)(a^2 + ab + b^2)\]

let \(a=(u+h)^{1/3}\) and \(b=u^{1/3}\)

you then get

\[(u+h)-u=[(u+h)^{1/3}-u^{1/3}][(u+h)^{2/3}+(u+h)^{1/3}u^{1/3}+u^{2/3}]\]

so

\[h=[(u+h)^{1/3}-u^{1/3}][(u+h)^{2/3}+(u+h)^{1/3}u^{1/3}+u^{2/3}]\]

Answer 27

awesome

Answer 28

oh my god, thank you so much, got it :D!
you guys are awesome ;)