# Finding the limit of a/(1-x^a) - b/(1-x^b) as x->1

1. Aug 6, 2005

### VietDao29

Can anyone help me understand this please?
This is one of the examples in the book I find really hard to understand.
$$\lim_{x \to 1} \left( \frac{a}{1 - x ^ a} - \frac{b}{1 - x ^ b} \right) \mbox{, ab \neq 0}$$
Here is what the book says:
In some neighbourhood of x = 1, $\alpha \neq 0$you have:
$$1 - x ^ \alpha = -\alpha (x - 1) - \frac{\alpha (\alpha - 1)}{2} (x - 1) ^ 2 + o((x - 1) ^ 2)$$
This is Taylor's series. I understand this.
And the book continues:
$$\frac{a}{1 - x ^ a} - \frac{b}{1 - x ^ b} = \frac{a(1 - x ^ b) - b(1 - x ^ a)}{(1 - x^a)(1 - x ^ b)} = \frac{a - b}{2} + o((x - 1) ^ 2)$$
This equation troubles me. I don't understand how they get it.
So they conclude:
$$\lim_{x \to 1} \left( \frac{a}{1 - x ^ a} - \frac{b}{1 - x ^ b} \right) = \frac{a - b}{2} \mbox{, ab \neq 0}$$
Any help would be appreciated.
Viet Dao,

2. Aug 7, 2005

### HallsofIvy

Staff Emeritus
Cute little problem! Look at the numerator and denominator of $$\frac{a(1-x^b)-b(1-x^a)}{(1-x^a)(1-x^b)}$$ separately. (Of course, that fraction is got by getting the common denominator. I assume it is the next equality that is bothering you.)

Using the approximation given, the numerator is $$(-ab(x-1)+\frac{ab(b-1)}{2}(x-1)^2)+ab(x-1)-\frac{ab(a-1)}{2}(x-1)^2$$ where I have simply not written the terms of degree higher than 2. Do you notice that, ignoring terms of power higher than 2, that reduces to one term?

The denominator is the product $$(-a(x-1)+\frac{a(a-1)}{2}(x-1)^2)(-b(x-1)+\frac{b(b-1)}{2}(x-1)^2)$$. Again, I have dropped terms of power greater than 2. Also in multiplying that out, you should drop terms of power greater than 2. It reduces very nicely!

Last edited: Aug 7, 2005
3. Aug 7, 2005

### VietDao29

Thanks for the help, I get it now.
Viet Dao,