lim as x --> 1 of

(x^(1/3) - 1) / (x^(1/2) - 1)

The answer is supposed to be 2/3.

I tried to use conjugates, but I still end up with a 0 in the denominator.

Could someone work through it? Thanks!

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- Sep 19th 2013, 09:31 PMIdentityProblemHelp Finding a Limit
**lim as x --> 1 of**

(x^(1/3) - 1) / (x^(1/2) - 1)

The answer is supposed to be 2/3.

I tried to use conjugates, but I still end up with a 0 in the denominator.

Could someone work through it? Thanks! - Sep 19th 2013, 09:47 PMibduttRe: Help Finding a Limit
- Sep 19th 2013, 09:48 PMCesc1Re: Help Finding a Limit
Notice that you can express that as

lim as x--->1 of

f(x)/g(x)

where f(x)= x^(1/3)-1, and g(x)=x^(1/2)-1

Since both limits approach to 0, you can use L'Hopital's rule:

lim as x---> c of

f(x)/g(x)= lim as x-->c of

f'(x)/g'(x) - Sep 20th 2013, 03:37 PMSorobanRe: Help Finding a Limit
Hello, IdentityProblem!

Quote:

$\displaystyle \lim_{x\to1}\frac{x^{\frac{1}{3}} - 1}{x^{\frac{1}{2}} - 1} $

The conjugate of $\displaystyle (x^{\frac{1}{2}}-1)$ is $\displaystyle (x^{\frac{1}{2}}+1)$

The conjugate of $\displaystyle (x^{\frac{1}{3}}-1)$ is $\displaystyle (x^{\frac{2}{3}}+x^{\frac{1}{3}} + 1)$

Now try it again . . .

- Sep 21st 2013, 01:31 PMIdentityProblemRe: Help Finding a Limit
Thank you, Soroban!

That raises a new question: how do you find a conjugate from scratch? - Sep 21st 2013, 02:03 PMFelixFelicis28Re: Help Finding a Limit
I'm not exactly sure what you mean as a conjugate in this sense, but if you're asking how Soroban found the conjugate of $\displaystyle x^{\frac{1}{3}} - 1$ (what I'm going to presume is an expression $\displaystyle y$ such that $\displaystyle (x^{\frac{1}{3}} - 1) y = x - 1$) then this is a useful trick to have in mind:

$\displaystyle x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + \cdots + x y^{n-2} + y^{n-1}) \\ \implies x - 1 = (x-y) (x^{n-1} + x^{n-2} y + \cdots + xy^{n-2} + y^{n-1}) \bigg|_{n = 3, \ x\mapsto x^{\frac{1}{3}}, \ y = 1$