# Show continuity using analitycal definition

I want to prove that the function in $\mathbb{R}^+$:

$$f(x)=x^{1-\alpha}-x$$

for $\alpha\in(0,1)$, is continous using the "epsilon-delta" definition of limit. To do this, I try to show that $\lim_{x\to x_0}f(x)=f(x_0)=x_0^{1-\alpha}-x_0$ in the following way:

For all $\epsilon>0$ there exists a $\delta>0$ such that if $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<\epsilon$. Trying to find this $\delta$ I have manipulated this former expession to obtain: $$|f(x)-f(x_0)|=|x^{1-\alpha}-x_0^{1-\alpha}-(x-x_0)|$$ $$\leq |x^{1-\alpha}-x_0^{1-\alpha}|+|x-x_0|$$ $$\leq |x^{1-\alpha}-x_0^{1-\alpha}|+\delta$$

but I am not sure on how to further bound $|x^{1-\alpha}-x_0^{1-\alpha}|$. Any advice? Thanks a lot!

• For a real exponent $1-\alpha$, the continuity of $x \mapsto x^{1-\alpha}$ is somehow granted by definition. How do you define a power with real exponent? – Siminore Oct 31 '17 at 15:34
• What's the domain of f? – Aaron Maroja Oct 31 '17 at 15:37
• @AaronMaroja $\mathbb{R}^+$ – Weierstraß Ramirez Oct 31 '17 at 15:44

By factoring $x_0>0$, it is enough to analyze the continuity at $x=1$. So you want to work with $|x^{1-\alpha}-1|$ for $x$ close to $1$ (this is not essential, but it simplifies any computation a little bit).
I don't think you can get this bounded by (a multiple of) $|x-1|$ by algebraic methods. A key issue is how you even define $x^{1-\alpha}$. The standard way is to define $$x^{1-\alpha}=e^{(1-\alpha)\log x}.$$ The continuity then follows from the continuity of the logarithm and the exponential. We can get concrete estimates this way: if you write $x=1+d$, with $d$ small, then \begin{align} x^{1-\alpha}-1&=e^{(1-\alpha)\log (1+d)}-1=e^{(1-\alpha)(d-o(d^2))}-1 =(1-\alpha)(d-o(d^2))+o(d^2)\\ \ \\&=(1-\alpha)d+o(d^2). \end{align} But getting to this level of explicit estimates requires dealing with Taylor expansions (or, at least in this case, with the Mean Value Theorem), while the continuity of $\log$ and $\exp$ is a more basic fact.