Limit of rational function to rational power

1. Oct 22, 2009

vertciel

1. The problem statement, all variables and given/known data

Evaluate the limit, WITHOUT using l'Hôpital's rule:

$$\lim_{x \rightarrow -1} \frac{x^{1/3} + 1}{x^{1/5} + 1}$$

2. Relevant equations

3. The attempt at a solution

I tried to use the conjugate method which does not produce a useful outcome:

$$\underset{x\to -1}{\mathop{\lim }}\,\frac{{{x}^{1/3}}+1}{{{x}^{1/5}}+1}\left( \frac{{{x}^{1/3}}-1}{{{x}^{1/3}}-1} \right)=\underset{x\to -1}{\mathop{\lim }}\,\frac{{{x}^{2/3}}-1}{{{x}^{8/15}}-{{x}^{1/5}}+{{x}^{1/3}}-1}$$

2. Oct 22, 2009

Bohrok

$$u^{15} = x, ~then ~u^3 = x^\frac{1}{5} ~and ~u^5 = x^\frac{1}{3}$$

Then factor and cancel.

3. Oct 23, 2009

HallsofIvy

That's not the "conjugate method". $x^a- 1$ is not always "conjugate" to $x^a+ 1$. A and B (both including roots) are "conjugate" if and only if AB does not have any roots. $x^{1/2}- 1$ is conjugate to $x^{1/2}+ 1$ because $(x^{1/2}- 1)(x^{1/2}+ 1)= (x^{1/2})^2- 1^2= x- 1$ but $x^{1/3}+ 1$ is conjugate to $x^{2/3}- x^{1/3}+ 1$ because $(x^{1/3}+ 1)(x^{2/3}- x^{1/3}+ 1)= x+ 1$

Instead use the facts that $x+ 1= (x^{1/3}+ 1)(x^{2/3}- x^{1/3}+ 1)$ and that $x+ 1= (x^{1/5}+ 1)(x^{4/5}- x^{3/5}+ x^{2/5}- x^{1/5}+ 1)$. Those are the "conjugates".

4. Oct 23, 2009

vertciel

Thank you for your responses, Bohrok and HallsofIvy.

@HallsofIvy: Thanks for your clarification on conjugates.

How would you define a real conjugate then? Are two terms $$x$$ and $$y$$ conjugates of each other if and only $$x \times y$$ are of degree 1 and do not have any fractional exponents?

5. Oct 23, 2009

HallsofIvy

Not necessarily of degree one but "does not have fractional exponents", yes.