1. Jun 1, 2014

Cardinality

The problem: (See attachment)

I rationalize the numerator and then simplify to get:

lim (3(1+sqrtx) - 2(1+cubert(x)))/(1-x)
x->1

However, I do not know what to do after this point. If someone could please help me out, that would be great!

Thank you!

P.S: Look carefully at the attachment, and note that the second denominator has x^(1/3).

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• Limit Problem.PNG
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Last edited: Jun 1, 2014
2. Jun 1, 2014

pasmith

It may be simplest to set $y = x^{1/6}$. Then $$\frac{3}{1 - x^{1/2}} - \frac{2}{1 - x^{1/3}} = \frac{3}{1 - y^3} - \frac{2}{1 - y^2}$$ and now you can use $$1 - y^3 = (1 - y)(1 + y + y^2) \\ 1 - y^2 = (1 - y)(1 + y)$$ before putting everything on a common denominator.

3. Jun 1, 2014

Cardinality

I tried what you recommended but I end off with:

lim x->1 (3(1+y) - 2(1+y+y^2))/((1+y+y^2)(1+y))

Note: I know the final answer is supposed to be 1/2.

4. Jun 1, 2014

Ray Vickson

The way I like to do problems like this one is to put $x = 1+r$ and then expand $\sqrt{x} = (1+r)^{1/2}$ and $x^{1/3} = (1+r)^{1/3}$ in powers of $r$, for small $r$.

5. Jun 1, 2014

Cardinality

I'm sorry, I don't think I quite understand Ray Vickson.

If anyone is willing to explain their method a little more in-depth it would be greatly appreciated!

6. Jun 1, 2014

Ray Vickson

Basically, if $x \to 1$ then when we write $x = 1 + r$ we are taking $r \to 0$; in other words, $r = x-1 \to 0$. If you know how to expand $(1+r)^{1/2}$ and $(1+r)^{1/3}$ in powers of $r$ for small $r$, then just go ahead and do it to see what happens. If you don't know how to do those expansions, then my suggestion will not be of much use to you.

Anyway: your expression is incorrect. When starting from the expression in the attachment and then rationalizing the denominators, you do not get the function you wrote. I suggest you go back to square one.

Last edited: Jun 1, 2014
7. Jun 1, 2014

8. Jun 1, 2014

Cardinality

Thanks Ray, but I unfortunately do not know how to expand. If there is another method, please suggest so!

LCKurtz, think you could help? Sorta stealing my thread here aha

9. Jun 1, 2014

mafagafo

I confirm this. Just redo it.

10. Jun 1, 2014

Cardinality

11. Jun 1, 2014

mafagafo

You misunderstood. The FIRST one is RIGHT. (IT IS SOLVABLE).
BUT, when you told us that:
(3(1+sqrtx) - 2(1+cubert(x)))/(1-x) = (3(1+y) - 2(1+y+y^2))/((1+y+y^2)(1+y)), if y = x ^ (1/6)
You can easily check this numerically, but just redo your algebra. If you can't find your mistake, post your step-by-step attempt and we will find it (or try really hard to).

12. Jun 1, 2014

Cardinality

Oh ok.

Here are my attempts at the solution (see attachments). The attachments are two separate work-in-progress solutions with slightly varying steps. If someone could please help me out that'd be great. This assignment is due tomorrow morning at 8 for me (it's almost midnight right now).

This is the only question that really gave me trouble and I'm at ends wit right now from constantly attempting this.

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• Try 2.jpg
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Last edited: Jun 1, 2014
13. Jun 1, 2014

SammyS

Staff Emeritus

It appears that you handled the first term well, correctly rationalizing the denominator (not the numerator). It helps to show your steps if we are to give good suggestions.)
3/(1-√x) = 3((1+√x)/((1-√x)(1+√x)) = 3((1+√x)/(1-x)​

Of course, this works because a difference times sum a gives the difference of squares.

$(1 - y)(1 + y) = 1 - y^2$

So if y = √x , then 1 - y2 is 1 - x .

However, this does no good for a cube root.

pasmith showed how to factor a difference of cubes.
$(1 - u)(1 + u + u^2) = 1 - u^3$​

What is this if $u =\sqrt[3]{x}\ ?$

14. Jun 1, 2014

Cardinality

I posted my work in the previous post please take a look!

15. Jun 1, 2014

SammyS

Staff Emeritus
The second sheet looks more promising.

Now factor $\ 2y^2-y-1\ .\$

There was no need to expand the denominator by multiplying it out.

16. Jun 1, 2014

Cardinality

Factor out from where? The main body so it's the numerator to (1-y)?
Where would I go from there?

17. Jun 1, 2014

SammyS

Staff Emeritus
Factor $\ 2y^2-y-1\$ period .

Not from anything.

Just factor it into the product of two binomials.

18. Jun 1, 2014

Cardinality

Edit: Yeah I'm just gonna hope you didn't see this post...my god, that was horrible. LOL

Last edited: Jun 1, 2014
19. Jun 1, 2014

Cardinality

Two things:

1). I'm an idiot.

2). I love you guys, thank you. Did I mention I'm an idiot?

Note to self: Move your *** from your desk from time to time and get a breather...or you forget to factor

Thanks again, & good night friends! :zzz: