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