\begin{align} \frac{\mathrm{d}}{\mathrm{d} x}\ln x &= \lim_{h\to0} \frac{\ln(x + h) - \ln(x)}{h} \\ &= \lim_{h\to0} \frac{\ln(\frac{x + h}{x})}{h} \\ &= \lim_{h\to0} \frac{\ln(1 + \frac{h}{x})}{h} \\ \end{align}

\begin{align} &= \lim_{h\to0}{\ln(1 + \frac{h}{x})}^\frac{1}{h} \\ \end{align} Now this is the part I'm asking about, because I see most people set the reciprocal of my limit, but this one seems to work: \begin{align} n=\frac{x}{h}, h=\frac{x}{n}, \frac{1}{h} = \frac{n}{1}*\frac{1}{x} \end{align} So when h tends towards 0, n tends towards infinity. \begin{align} &= \lim_{n\to\infty}{\ln\biggl((1 +\frac{1}{n})}^n\biggl)^\frac{1}{x} \\ \end{align} Here we have \begin{align} {\ln\lim_{n\to\infty}\biggl((1 +\frac{1}{n})}^n\biggl)^\frac{1}{x} \\ \end{align} \begin{align} =\frac1x\ln(e) = \frac1x \end{align} This is my first post, I was just curious. Please don't berate me if it's a silly question.

  • $\begingroup$ When considering the limit $h \to 0$, I think it's necessary to take both sides to show that the limit exists. Maybe add: $\lim_{h \to 0^- } = -\infty$? Proof still works though $\endgroup$
    – John Lou
    Apr 6, 2018 at 15:15
  • $\begingroup$ It seems to be right, but when you changed from h to n, at first you still wrote "as h approaches 0" under the limit, you should write "as n approaches infinity" there because you changed the variable. And one more thing, if you want your proof to be completely rigorous, you need to verify your change of variables in limit in some way $\endgroup$ Apr 6, 2018 at 15:16

2 Answers 2


I think this is OK. You should acknowledge that at one point you are using the fact that the function raising something to a fixed power is continuous: $$ \lim (\text{something}^a)= (\lim (\text{something}) )^a . $$

Moreover, the usual definition of exponentiation for an arbitrary exponent $a$ depends on the natural logarithm: $$ \text{something}^a = \exp(a (\ln \text{something})) $$ so there's a lot of serious analysis (sometimes skipped over the first time through in calculus) in the correct algebraic manipulations you're doing.

That said, this is a strange exercise. Presumably you have defined $\ln$ as the inverse of exponentiation, so that $$ \exp(\ln(x)) = x . $$ Then the formula for the derivative of $\ln$ follows from the chain rule.

  • $\begingroup$ Ok, thanks. I did the proof in this way because I have yet to learn the chain rule, but I will be very soon :) $\endgroup$
    – DevinJC
    Apr 6, 2018 at 15:25
  • 3
    $\begingroup$ Should also mention that ln is continuos $\endgroup$
    – Ant
    Apr 6, 2018 at 16:29

yeah your answer looks correct , I'd probably do it this way tho ;

$\lim_{h\to0}\frac1h\ln\big(1+\frac hx) \\= \lim_{h\to0}\frac xh \frac1x\ln\big(1+\frac hx) \\= \frac1x\lim_{h\to0}\ln\big(1+\frac 1{\frac xh})^\frac xh\\ = \frac1x\ln(e) \\= \frac1x$


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