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How do I show that $$\lim_{x \to x_0} \frac{x^{\alpha }-x_{0}^{\alpha } }{x-x_0} = \lim_{x \to 1} \frac{x_{0}^{\alpha }x^{\alpha }-x_{0}^{\alpha } }{x_0 x-x_0}$$

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For $x_0\neq 0$, let $x=yx_0$ with $y\to 1$

$$\lim_{x \to x_0} \frac{x^{\alpha }-x_{0}^{\alpha } }{x-x_0} = \lim_{y \to 1} \frac{(yx_0)^{\alpha }-x_{0}^{\alpha } }{(yx_0)-x_0} =\lim_{y \to 1} \frac{x_{0}^{\alpha }y^{\alpha }-x_{0}^{\alpha } }{x_0 y-x_0}$$

For $x_0=0$

$$\lim_{x \to x_0} \frac{x^{\alpha }-x_{0}^{\alpha } }{x-x_0} =\lim_{x \to 0} x^{\alpha-1 }$$

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  • $\begingroup$ In the case $x_0 = 0$ your starting equation $x = yx_0$ does not define $y$. You should at least point out that the case $x_0 = 0$ must be treated separately. $\endgroup$ – Gibbs Apr 14 '18 at 9:50
  • $\begingroup$ @Gibbs yes of course the case $x_0=0$ should be treated a part $\endgroup$ – gimusi Apr 14 '18 at 9:54
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Just define $y = \frac{x}{x_0}$ and observe that when $x$ goes to $x_0$, $y$ goes to 1. Then you can rewrite $$\lim_{x \rightarrow x_0} \frac{x^a-x_0^a}{x-x_0} = \lim_{y\rightarrow 1} \frac{y^ax_0^a-x_0}{yx_0-x_0}. $$ I divided by $x_0$, so treat the case $x_0 = 0$ separately.

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