I am re-reading my calculus book and it has the following proof for the derivative of $\log_ax$:

$$(\log_ax)'=\lim_{h\to 0}\frac{{\log_a{(x+h)}}-\log_a{x}}{h}=\lim_{h\to0}\frac{\log_a{\left(\frac{x+h}{x}\right)}}{h}$$

$$ =\frac{1}{x}\lim_{h\to0}{\left[\log_a{\left(1+\frac{h}{x}\right)^{\frac{x}{h}}}\right]} $$

That is all well and good, but I don't get how they make the next step:

$$ =\frac{1}{x}\log_a{\left[\lim_{h\to0}{\left(1+\frac{h}{x}\right)^{\frac{x}{h}}}\right]} $$

I could only find different (and simpler) proofs online, and this question is just about how that last step was done. Thank you.

up vote 3 down vote accepted

For continuous function, we can interchange the limit with the function.

That is if a function $f$ is continuous at $x_0$, then $$\lim_{x \to x_0} f(x) = f\left(\lim_{x \to x_0} x\right)=f(x_0)$$

The last move is valid as $\log_a(.)$ is a continuous function.

  • +1, never knew this! Follow-up question (not OP): If $f_1, f_2, \dots, f_n: \mathbb{R} \to \mathbb{R}$ are continuous functions at $x_0$, then would $$\lim_{x\to x_0}(f_n(\dots(f_2(f_1(x)))\dots)) = f_n(\dots(f_2(f_1(\lim_{x\to x_0}x)))\dots) \text{?}$$ Would this hold for infinitely many functions, or would we get tied up in complications involving AOC? – Andrew Tawfeek Sep 4 '17 at 6:03
  • 2
    composition of finitely many continuous function is continuous, hence yes. It is possible to compose a continuous function infinitely many times such that the limit exists and yet it is not continuous. – Siong Thye Goh Sep 4 '17 at 6:16
  • That's interesting! Do you know where I could read up more about this? Is it something that's studied in Real Analysis? – Andrew Tawfeek Sep 4 '17 at 6:21
  • Thank you very much. I will read more about this. – sempiedram Sep 4 '17 at 6:26
  • 2
    @AndrewTawfeek result about finite composition is typically covered in a calculus/ analysis class. Infinite composition I guess is what I learned by spending time on this site. – Siong Thye Goh Sep 4 '17 at 18:27

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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