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

#### StereoBucket

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.
$$\displaystyle \lim_{x\to\1} \frac{x^a - 1}{x-1}$$ $$\displaystyle , a\in\mathbb{R} , a \geq 0$$

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

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#### StereoBucket

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.

#### 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 $$\displaystyle t=0$$.

• StereoBucket

#### Plato

MHF Helper
$$\displaystyle \frac{x^a-1}{x-1}=\frac{e^{a\log x}-1}{e^{\log x}-1}=\frac{e^{ah}-1}{h}\cdot\frac{h}{e^h-1}$$
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?

#### Archie

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