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


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!

  • $\begingroup$ 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? $\endgroup$ – Siminore Oct 31 '17 at 15:34
  • $\begingroup$ What's the domain of f? $\endgroup$ – Aaron Maroja Oct 31 '17 at 15:37
  • $\begingroup$ @AaronMaroja $\mathbb{R}^+$ $\endgroup$ – 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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.