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

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In summary, the author finds a problem with an equation that reduces to one term when simplifying the denominator.
  • #1
VietDao29
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Can anyone help me understand this please?
This is one of the examples in the book I find really hard to understand.
[tex]\lim_{x \to 1} \left( \frac{a}{1 - x ^ a} - \frac{b}{1 - x ^ b} \right) \mbox{, ab \neq 0}[/tex]
Here is what the book says:
In some neighbourhood of x = 1, [itex]\alpha \neq 0[/itex]you have:
[tex]1 - x ^ \alpha = -\alpha (x - 1) - \frac{\alpha (\alpha - 1)}{2} (x - 1) ^ 2 + o((x - 1) ^ 2)[/tex]
This is Taylor's series. I understand this.
And the book continues:
[tex]\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)[/tex] :confused:
This equation troubles me. I don't understand how they get it.
So they conclude:
[tex]\lim_{x \to 1} \left( \frac{a}{1 - x ^ a} - \frac{b}{1 - x ^ b} \right) = \frac{a - b}{2} \mbox{, ab \neq 0}[/tex]
Any help would be appreciated.
Viet Dao,
 
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  • #2
Cute little problem! Look at the numerator and denominator of [tex]\frac{a(1-x^b)-b(1-x^a)}{(1-x^a)(1-x^b)}[/tex] 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 [tex](-ab(x-1)+\frac{ab(b-1)}{2}(x-1)^2)+ab(x-1)-\frac{ab(a-1)}{2}(x-1)^2[/tex] 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 [tex](-a(x-1)+\frac{a(a-1)}{2}(x-1)^2)(-b(x-1)+\frac{b(b-1)}{2}(x-1)^2)[/tex]. 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!
 
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  • #3
Thanks for the help, I get it now. :smile:
Viet Dao,
 

1. What is the limit of the function as x approaches 1?

The limit of this function as x approaches 1 is indeterminate. This means that the limit does not exist or it cannot be determined using traditional methods.

2. How can I find the limit of this function?

To find the limit of this function, you can use L'Hopital's rule or try to simplify the expression by factoring or canceling out common terms. You can also use a graphing calculator or a graphing software to visualize the behavior of the function as x approaches 1.

3. Why is the limit of this function indeterminate?

The limit of this function is indeterminate because as x approaches 1, the function becomes undefined. This is because both the numerator and denominator become 0, resulting in an indeterminate form of 0/0.

4. Is there a specific value that can be assigned to the limit of this function?

No, there is not a specific value that can be assigned to the limit of this function as x approaches 1. This is because the limit is indeterminate, meaning it cannot be determined using traditional methods.

5. Can the limit of this function be calculated for values of x other than 1?

Yes, the limit of this function can be calculated for values of x other than 1. However, the value of the limit may be different depending on the value of x. This is because the behavior of the function may change as x approaches a different value.

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