I have encountered the following proof of derivative from limits: let $f(x)= \frac{\ln\ x}{x}$

\begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{\frac{\ln(x+\Delta x)}{x+\Delta x} - \frac{\ln\ x}{x}}{\Delta x} \end{equation}

\begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{x\ ln(1+\frac{\Delta x}{x}) - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)} \end{equation}

\begin{equation} f'(x) = lim_{\Delta x \rightarrow 0} \frac{\Delta x - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)} \end{equation}

\begin{equation} f'(x) = \frac{1-ln\ x}{x^2} \end{equation}

From the 2nd to the 3rd line, I don't understand how the term $x\ ln(1+\frac{\Delta x}{x})$ was simplified to $\Delta x$ ?


Hint: $$\ln (1+x) \sim x, x \to 0$$


Let's consider $\lim_\limits{\Delta x \to 0} \frac{x\ ln(1+\frac{\Delta x}{x}) - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}$ and divide it in 2 parts:

  1. $\lim_\limits{\Delta x \to 0} \frac{x\ ln(1+\frac{\Delta x}{x}) }{x\Delta x(x+\Delta x)}= \lim_\limits{ \Delta x \to 0}\left[\frac{xln(1+\frac{\Delta x}{x})}{\Delta x}\cdot \frac{1}{x(x+\Delta x) }\right]=\\ =\lim_\limits{ \Delta x \to 0} \frac{xln(1+\frac{\Delta x}{x})}{\Delta x} \cdot \lim_\limits{ \Delta x \to 0} \frac{1}{x(x+\Delta x) }= \frac{1}{x^2}$

  2. $\lim_\limits{\Delta x \to 0} \frac{\Delta x\ ln\ x}{x\Delta x(x+\Delta x)}=\frac{\ln x}{x^2}$

All steps are exact and based on well known theorems.

And separately about limit $\lim_\limits{ \Delta x \to 0} \frac{xln(1+\frac{\Delta x}{x})}{\Delta x} = \lim_\limits{ \Delta x \to 0} \frac{ln(1+\frac{\Delta x}{x})}{\frac{\Delta x}{x}} =\lim\limits_{t\to 0} \frac{\ln (1+t)}{t}=1 $: here is applied a theorem about limit of functions composition i.e. changing variable, which is standard theorem.

Let me say, that it is important to study and remember, that in multiplication it is possible to change sub-expression with its equivalence expression, if all appropriate conditions hold: assume $f \sim \phi$ i.e. $\lim \frac {f}{\phi} = 1$. Then $\lim (f \cdot g)= \lim \left(\dfrac {f}{\phi} \cdot \phi \cdot g \right) = \lim \frac {f}{\phi} \cdot \lim (\phi \cdot g) = \lim (\phi \cdot g)$ when all written limits exists.

  • $\begingroup$ I see, thanks! I didn't think of this approximation $\endgroup$ – Tianxun Zhou Mar 12 at 2:56
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    $\begingroup$ This does not mean that we can replace $\log(1+x)$ by $x$ during the limit evaluation process. It's a common mistake committed by many. $\endgroup$ – Paramanand Singh Mar 12 at 2:59
  • $\begingroup$ @TianxunZhou: please understand that limit evaluation is not based on the use of approximations, but it is done by using various theorems and formulas related to evaluation of limits. $\endgroup$ – Paramanand Singh Mar 12 at 3:05
  • $\begingroup$ @Paramanand Singh. It is hint, you should elaborate it: consider $t=\Delta x/x$ and having $t \to 0$ you can use it for $\ln (1+ \Delta x/x)$. The mistake is the naming mistake of what did not understand. This is a standard method/theorem about changing a variable in the limit and applies even to non-continuous functions in appropriate conditions. $\endgroup$ – zkutch Mar 12 at 3:10
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    $\begingroup$ Here is an example when you will see the mistake and accept it. Replace $\log(1+x)$ by $x$ in the evaluation of limit of $(\log(1+x)-x)/x^2$ as $x\to 0$ to get $0$. The correct answer is $-1/2$. $\endgroup$ – Paramanand Singh Mar 12 at 3:17

That step is wrong. You can't replace a sub-expression by another unless they are equal. Any appeal to the use of standard limits in this manner is a complete disregard for limit laws. If it's coming from a textbook then shame!!

The correct way to handle this is as follows (using $h$ in place of $\Delta x$ to reduce typing effort) \begin{align} f'(x) &=\lim_{h\to 0}\frac{x\log(1+(h/x))-h\log x} {xh(x+h)} \notag\\ &=\lim_{h\to 0}\frac{\log(1+(h/x)) }{h/x} \cdot\frac{1}{x(x+h)}-\frac{\log x} {x(x+h)} \notag\\ &=1\cdot\frac{1}{x\cdot x} - \frac{\log x} {x\cdot x} \notag\\ &=\frac{1-\log x} {x^2}\notag \end{align}

  • $\begingroup$ Thanks for your answer, $\lim_{\Delta x \to 0} \frac{log(1+h/x)}{h/x} = 1$ is obtained through L'hopital's rule am I right? $\endgroup$ – Tianxun Zhou Mar 12 at 3:09
  • $\begingroup$ @TianxunZhou: the result $\lim_{t\to 0}\dfrac{\log(1+t)}{t}=1$ is a standard property of logarithm function and it can be proved using the definition of logarithm. Use of L'Hospital's Rule would be circular here. $\endgroup$ – Paramanand Singh Mar 12 at 3:22

Using $h$ in place of $\Delta x$
$\begin{align} f'(x) &=\lim_{h\to 0}\frac{x\log(1+(h/x))-h\log x} {xh(x+h)} \notag\\ \end{align}$
Use series expansion of $\log (1+h/x)$, where $|h/x|\lt 1$,
$\log(1+h/x)=\frac hx-\frac 12(\frac hx)^2+\frac 13(\frac hx)^3-\cdots$
$\begin{align} f'(x)&=\lim_{h\to 0}\frac{h(1 +\text{higher powers of $h$)}-h\log x}{xh(h+x)}\\ &= \lim_{h\to 0}\frac{(1+\text{higher powers of $h$})-\log x}{x(h+x)}\\ &=\frac{1-\log x}{x^2}\end{align}$

  • $\begingroup$ It is best to write $\log(1+h/x)=h/x+o(h/x)$ instead of using a series. This is called little o notation. $\endgroup$ – Paramanand Singh Mar 12 at 4:45
  • $\begingroup$ @ParamanandSingh: Thanks a lot for suggestion. I get confused between Big $O$ and little $o$ so I tend to avoid it. You may please suggest some source of Big O and little o so that I can learn it. $\endgroup$ – Koro Mar 12 at 4:46
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    $\begingroup$ This is not a big deal that you need some specific source to learn it. I find little o more convenient. For me the notation $o(\text{something}) $ denotes an expression (without explicitly writing that expression) which when divided by that "something" tends to $0$. In symbols, by definition, $o(f(x)) /f(x) \to 0$. $\endgroup$ – Paramanand Singh Mar 12 at 4:51
  • $\begingroup$ @ParamanandSingh: I think $o(f(x))/f(x)\to 0$ as $f(x)\to 0$. Anyways, I'll have a look into it. Thanks. $\endgroup$ – Koro Mar 12 at 4:54
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    $\begingroup$ That's not the case always. For example if $n\to\infty $ we can write $\sqrt {n} =o(n) $. $\endgroup$ – Paramanand Singh Mar 12 at 4:57

Hint Using the power law of logarithm and definition of e, we can change 2nd line to $\to$

$f'(x) = \lim_{\Delta x \to 0} \frac{x\ ln(1+\frac{\Delta x}{x}) - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}$

$=\lim_{\Delta x \to 0} \frac{\Delta x\ ln(1+\frac{\Delta x}{x})^{\frac{x}{ \Delta X}}}{x\Delta x(x+\Delta x)}-\lim_{\Delta x \to 0} \frac{ \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}$

$=\frac{\ \lim_{\Delta x \to 0} ln(1+\frac{\Delta x}{x})^{\frac{x}{ \Delta X}}}{\lim_{\Delta x \to 0} x(x+\Delta x)}-\lim_{\Delta x \to 0} \frac{ \ ln\ x}{x(x+\Delta x)} $

$=\frac{ln(\ \lim_{\Delta x \to 0} (1+\frac{\Delta x}{x})^{\frac{x}{ \Delta X}})}{\lim_{\Delta x \to 0} x(x+\Delta x)}-\lim_{\Delta x \to 0} \frac{ \ ln\ x}{x(x+\Delta x)} $ limit property of continuous function

$=\frac{\ln e}{\lim_{\Delta x \to 0} x(x+\Delta x)}-\lim_{\Delta x \to 0} \frac{ \ ln\ x}{x(x+\Delta x)} $

With these intermediate steps and definition of e , we can go from 2nd line to 3rd line .

  • $\begingroup$ One can't replace that sub-expression by its limit $e$. $\endgroup$ – Paramanand Singh Mar 12 at 3:02
  • $\begingroup$ @Singh Could you please elaborate why the limit theorem does not allow this? $\endgroup$ – BStar Mar 12 at 4:09
  • $\begingroup$ Well because there is no such theorem (algebra of limits in particular) which allows you to do this. You may find some examples where it does not work. In case you are aware of some theorem which allows it please mention that. $\endgroup$ – Paramanand Singh Mar 12 at 4:43
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    $\begingroup$ @koro thanks for proving the change of limit variable, just taking it for granted in my answer -:). $\endgroup$ – BStar Mar 12 at 6:10
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    $\begingroup$ @Koro: I have discussed this topic at length in this answer and a formal discussion with proofs is available here. $\endgroup$ – Paramanand Singh Mar 12 at 9:30

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