# Help Finding a Limit

• Sep 19th 2013, 10:31 PM
IdentityProblem
Help 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, 10:47 PM
ibdutt
Re: Help Finding a Limit
• Sep 19th 2013, 10:48 PM
Cesc1
Re: 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, 04:37 PM
Soroban
Re: Help Finding a Limit
Hello, IdentityProblem!

Quote:

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

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

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

Now try it again . . .
• Sep 21st 2013, 02:31 PM
IdentityProblem
Re: Help Finding a Limit
Thank you, Soroban!

That raises a new question: how do you find a conjugate from scratch?
• Sep 21st 2013, 03:03 PM
FelixFelicis28
Re: Help Finding a Limit
Quote:

Originally Posted by IdentityProblem
Thank you, Soroban!

That raises a new question: how do you find a conjugate from scratch?

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 $x^{\frac{1}{3}} - 1$ (what I'm going to presume is an expression $y$ such that $(x^{\frac{1}{3}} - 1) y = x - 1$) then this is a useful trick to have in mind:

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