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

Jan 2017
6
3
Banja Luka
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.
 
Last edited:
Jan 2017
6
3
Banja Luka
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.
 
Dec 2013
2,002
757
Colombia
\(\displaystyle \frac{x^a-1}{x-1}=\frac{e^{a\log x}-1}{e^{\log x}-1}=\frac{e^{ah}-1}{h}\cdot\frac{h}{e^h-1}\)
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 \(\displaystyle t=0\).
 
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Plato

MHF Helper
Aug 2006
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\(\displaystyle \frac{x^a-1}{x-1}=\frac{e^{a\log x}-1}{e^{\log x}-1}=\frac{e^{ah}-1}{h}\cdot\frac{h}{e^h-1}\)
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?
 
Dec 2013
2,002
757
Colombia
I haven't taken a derivative, I've recognised a difference quotient.