Question

A limit problem.

Answer 1

@mebs Is this you throwing in a fun question or you really asking how to do this?

Answer 2

I am very serious got an exam on Monday buddy.

Answer 3

Oh. I rationalize the numerator.
There could be another way.

Answer 4

That was a first step i did.

Answer 5

I tried that I get infinity * 0

Answer 6

I took x3 t the roots so I had cube roots of (1+x2/x3) and (1-x2/x3)
e knowing x2/x3 <<1,
I took cube root as 1 + 1/3 (x2/x3) and 1 - 1/3 (x2/x3)

subtracting gave me x[2 x2/3 x3] which goes to 2/3 as x goes to infinity.

Answer 7

So you got something like this:
\[u^\frac{1}{3}-v^\frac{1}{3}=\frac{(u^\frac{1}{3}-v^\frac{1}{3})(u^\frac{2}{3}+v^\frac{1}{3}u^\frac{1}{3}+v^\frac{2}{3})}{(u^\frac{2}{3}+v^\frac{1}{3}u^\frac{1}{3}+v^\frac{2}{3})} \]

Answer 8

\[=\frac{u-v}{u^\frac{2}{3}+v^\frac{1}{3}u^\frac{1}{3}+v^\frac{2}{3}}\]

Answer 9

I don't recognize it you may be right, I worked with the cube root of (1 + x2/x3) etc.

Answer 10

Wow wait what was the factor rule you used for the first part... factoring an odd polynomial...

Answer 11

so you multiply by its reciprocal and factor why did you use u substitution its still x.

Answer 12

how did you do that that is like pretending that

its (a)^1/3-(b)^1/3

Answer 13

\[=\frac{(x^3+x^2)-(x^3-x^2)}{(x^3+x^2)^\frac{2}{3}+(x^3-x^2)^\frac{1}{3}(x^3+x^2)^\frac{1}{3}+(x^3-x^2)^\frac{2}{3}}\]

Answer 14

I was just putting u and v to make it look prettier

Answer 15

well one day I was like how do i rationalize a difference of cube roots
and i kept putting terms inside a little factor bubble and multiplying and seeing what i was missing and then putting that in and kept doing it until when i multiplied everything zeroed out except the difference of cube roots
i did this because i was trying to evaluate a limit
and i know rationalizing is an important tactic in evaluating limits

Answer 16

Ahhh that's great brilliant .. Thanks a lot !

Answer 17

however i wonder if there is an easier approach
the way i mention will work
but i had to do some more steps to get the answer

Answer 18

@mebs i think i'm onto something else
it might be prettier

Answer 19

im still here

Answer 20

Yeah that is the only way I can really think of. We could have factor out a x^(2/3) to start but that doesn't matter much. Or I can't make it matter much anyways. I'm saying the way I went out about you could have if you wanted.

Answer 21

Now looking at the bottom the highest exponent looks like x^2

Answer 22

So I divided both top and bottom by x^2

Answer 23

ok that makes sense

Answer 24

like we have these terms in the denominator (x^3+x^2)^(2/3)=x^3(2/3)+...blah,
(x^6-x^4)^(1/3)=x^2+blah,
and
(x^3-x^2)^(2/3)=x^2+blah
if you see what I mean

Answer 25

3(2/3)=2
6(1/3)=2

Answer 26

alright I guess that these powers are annoying but if we do divide we would have to change it .. alright that works

Answer 27

now you should see all three of your fractions in the denominator each go to one and your top should go to 2

Answer 28

yea it stays 2 at the bottom because of 2x^2

Answer 29

so the answer is 2!!

Answer 30

no

Answer 31

alright still listening.

Answer 32

2/3

Answer 33

could you do an example of applying that factoring thing you just did with like

x^(1/2)-y(1/2) step by step?

Answer 34

\[\lim_{x \rightarrow \infty} \frac{\frac{2x^2}{x^2}}{\frac{(x^3+x^2)^\frac{2}{3}}{x^2}+\frac{(x^6-x^4)^\frac{1}{3}}{x^2}+\frac{(x^3-x^2)^\frac{1}{3}}{x^2}}\]

Answer 35

and yeah sure

Answer 36

Oh I see its \[(1+\frac{1}{x}) + (1+\frac{1}{x})+ (1+\frac{1}{x})\]

Answer 37

and wait do you mean rationalize a difference of square roots?

Answer 38

so I think the general formula you just showed is \[x^n-a^n = (x-a)(x^{n-1} + ax^{n-2} ....a^{n-1)}\]

Answer 39

yea like pretend that we were doing that problem with some other variables a little more simpler ones that aren't fractions...

Answer 40

and yeah i'm pretty sure there is a formula for what i was doing
i bet i could find one by looking for patterns with fractional exponents

Answer 41

lets pretend I was factoring an odd functional exponent like \[t^{1/3}-r^{1/3}= \text{step by step rationalie and factor please}\]

Answer 42

Do you mean factor t-r with a difference of square roots, cube roots, 4th roots,...,nth roots

Answer 43

I meant what you did just now.

Answer 44

with that u^(1/3)-v^1/3

Answer 45

well me said \[u-v=(u^\frac{1}{3}-v^\frac{1}{3})(u^\frac{2}{3}+u^\frac{1}{3}v^\frac{1}{3}+v^\frac{2}{3})\]

Answer 46

i didn't mean to say me
that sounded weird

Answer 47

oHHHH I SEE IT OFMG yesss

Answer 48

all you did was divide both sides by that large term.