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When trying to calculate the deriviative of $ln(x)$, consider

$$ \lim\limits_{h \rightarrow 0} ln((1+\frac{h}{x})^{\frac{1}{h}})$$

Can I immediately deduce $$ ln(e^{\frac{1}{x}})= \frac{1}{x}$$

Or do I have to make the extra step of switching the lim and ln

$$\lim\limits_{h \rightarrow 0} ln((1+\frac{h}{x})^{\frac{1}{h}}) = ln\lim\limits_{h \rightarrow 0} ((1+\frac{h}{x})^{\frac{1}{h}})$$

This question is inspired by this comment https://math.stackexchange.com/a/1221383/360371 where the user said the switch of ln and lim is allowed, because ln(x) is continuous for $x>0$.

Can somebody elaborate, why it depends on the continuity?

I know it involves the fact, that the continuity guarantees that the limit is the same whatever side you approach it from, but I can't quite make the connection here.

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Jonathan
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  • This might be helpful: http://math.stackexchange.com/questions/18800/list-of-rules-and-laws-for-calculating-limits – Simply Beautiful Art Jan 05 '17 at 21:32
  • Remember that by definition of continuity, you can interchange the order of the functional operation with limit operation. In symbols we say $f$ is continuous at $a$ if $$\lim_{x \to a} f(x) =f(a) =f(\lim_{x\to a} x) $$ – Paramanand Singh Jan 06 '17 at 09:36

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One characterization of continuity of a function $f:\mathbb R \to \mathbb R$ in some $x_0 \in \mathbb R$ is that for each sequence $(a_n)_{n\in\mathbb N} \subseteq \mathbb R$ converging towards $x_0$ it holds that $$\lim_{n\to \infty}f(a_n) = f(\lim_{n\to \infty} a_n)= f(x_0).$$ This is equivalent to for example $\epsilon$-$\delta$ definition of continuity (proof can be found in every Calculus I script).

Staki42
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