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$\lim_{x\rightarrow 1}\frac{x^r -1}{x-1}$

r is Rational_number, with L Hospital rule is this easy to solve, but how could I solve it in another way?

sasasqt
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    Notice that this is exactly the derivative of $x^r$ evaluated in $x = 1$. – Desura Dec 17 '16 at 09:25
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    The mean value theorem for $f(x) = x^r$ gives you the existence of a $\xi$ in $(x,1)$ s.t. $$\frac{x^r-1}{x-1} = f'(\xi) = r\xi^{r-1}$$ For $x\to 1$ also $\xi \to 1$ hence $$\lim_{x\to 1} \frac{x^r-1}{x-1} = \lim_{\xi\to 1} r\xi^{r-1} = r$$ – Gono Dec 17 '16 at 09:26

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We know that $$x^r-1 = (x-1)(x^{r-1} + x^{r-2} +\cdots +1)$$ Thus, $$\lim_{x\to 1} \frac{x^r-1}{x-1} = \lim_{x\to 1} (x^{r-1} + x^{r-2} +\cdots +1) = r$$

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    what if if r <0? does it hold? – sasasqt Dec 17 '16 at 09:25
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    Even more interesting, perhaps: what if $;r;$ is not even an integer? – DonAntonio Dec 17 '16 at 09:47
  • Suppose that this limit exists. I can suppose that $r = \frac{a}{b} $ and $a, b \in \mathbb{N}^*$. Then I have $$\lim\limits_{x \to 1} \frac{x^{\frac{a}{b}} -1}{x-1} = \lim\limits_{x \to 1} \frac{x^{\frac{a}{b}} -1}{x-1} = \lim\limits_{x \to 1} \frac{(x^{\frac{1}{b}})^a -1}{(x^{\frac{1}{b}})^b-1}. $$ Put $t = x^{\frac{1}{b}}$ and I will do as Rohan. – ToThichToan Dec 17 '16 at 20:49