# Thread: Finding a limit without using L'Hopital's rule or Maclaurin Series

1. ## Finding a limit without using L'Hopital's rule or Maclaurin Series

I know what the answer is (according to all resources I checked, it should be a). However I have no clue how to get to it without using L'Hopital.
$\lim_{x\to\1} \frac{x^a - 1}{x-1}$ $, a\in\mathbb{R} , a \geq 0$

Edit: should've probably said that $a\in\mathbb{R}$, and is not just integers.

2. ## Re: Finding a limit without using L'Hopital's rule or Maclaurin Series

Thanks, but I'm pretty sure no one would accept this as an answer as it assumes that a is an integer, when it may not be.
Sorry for not stating that information. I edited the post to reflect on that.

3. ## Re: Finding a limit without using L'Hopital's rule or Maclaurin Series

$\frac{x^a-1}{x-1}=\frac{e^{a\log x}-1}{e^{\log x}-1}=\frac{e^{ah}-1}{h}\cdot\frac{h}{e^h-1}$
where $h=\log x \to 0$

When we now take the limit, the first term is the derivative of $e^{at}$ evaluated at $t=0$, while the second is the reciprocal of the derivative of $e^t$ evaluated at $t=0$.

4. ## Re: Finding a limit without using L'Hopital's rule or Maclaurin Series

Originally Posted by Archie
$\frac{x^a-1}{x-1}=\frac{e^{a\log x}-1}{e^{\log x}-1}=\frac{e^{ah}-1}{h}\cdot\frac{h}{e^h-1}$
where $h=\log x \to 0$

When we now take the limit, the first term is the derivative of $e^{at}$ evaluated at $t=0$, while the second is the reciprocal of the derivative of $e^t$ evaluated at $t=0$.
How is this not equivalent to using one of the methods we were asked to avoid?

5. ## Re: Finding a limit without using L'Hopital's rule or Maclaurin Series

I haven't taken a derivative, I've recognised a difference quotient.