# 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.
$\displaystyle $$\lim_{x\to\1} \frac{x^a - 1}{x-1}$$$ $\displaystyle ,$a\in$\mathbb{R}$ , a \geq 0 $Thanks in advance. Edit: should've probably said that$\displaystyle a\in$\mathbb{R}$ $, and is not just integers. 2. ## Re: Finding a limit without using L'Hopital's rule or Maclaurin Series Thanks, but I'm pretty sure no one would accept this as an answer as it assumes that a is an integer, when it may not be. Sorry for not stating that information. I edited the post to reflect on that. 3. ## Re: Finding a limit without using L'Hopital's rule or Maclaurin Series$\displaystyle \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}$where$\displaystyle h=\log x \to 0$When we now take the limit, the first term is the derivative of$\displaystyle e^{at}$evaluated at$\displaystyle t=0$, while the second is the reciprocal of the derivative of$\displaystyle e^t$evaluated at$\displaystyle t=0$. 4. ## Re: Finding a limit without using L'Hopital's rule or Maclaurin Series Originally Posted by Archie$\displaystyle \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}$where$\displaystyle h=\log x \to 0$When we now take the limit, the first term is the derivative of$\displaystyle e^{at}$evaluated at$\displaystyle t=0$, while the second is the reciprocal of the derivative of$\displaystyle e^t$evaluated at$\displaystyle 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.