1

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 $\Delta x$ ?

Rodrigo de Azevedo
  • 1
Tianxun Zhou
  • 161

4 Answers4

2

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

Addition.

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.

zkutch
  • 13,410
  • I see, thanks! I didn't think of this approximation – Tianxun Zhou Mar 12 '21 at 02:56
  • 2
    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. – Paramanand Singh Mar 12 '21 at 02:59
  • @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. – Paramanand Singh Mar 12 '21 at 03:05
  • @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. – zkutch Mar 12 '21 at 03:10
  • 1
    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 $-1/2$. – Paramanand Singh Mar 12 '21 at 03:17
  • That's one of the reasons we need rigor and standard theorems. Unfortunately in calculus whatever works is taken as a standard. Some of these techniques work by fluke. – Paramanand Singh Mar 12 '21 at 03:19
  • @Paramanand Singh. See addition. All steps are correct and there is place where hint is used. – zkutch Mar 12 '21 at 03:24
  • The limit evaluation of first part is having the same problem. Do not replace $\log(1+(\Delta x /x)) $ by $\Delta x/x$. This is not possible via any of theorems in limits. – Paramanand Singh Mar 12 '21 at 03:27
  • It is not problem, it is possible to change equivalence expression in multiplication, when limit of all parts exist. – zkutch Mar 12 '21 at 03:30
  • Ok, please mention in your post about equivalents and the laws of equivalents being used. A typical calculus may not be aware of these terms based on the country where they studied. – Paramanand Singh Mar 12 '21 at 03:34
  • @Paramanand Singh. Is it enough clear now? If not point exactly to place with doubt. – zkutch Mar 12 '21 at 03:43
  • +1 now. Also note that the way this has been done in question is not exactly about replacing equivalents in multiplication. – Paramanand Singh Mar 12 '21 at 04:41
1

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}

Paramanand Singh
  • 87,309
  • 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? – Tianxun Zhou Mar 12 '21 at 03:09
  • @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. – Paramanand Singh Mar 12 '21 at 03:22
0

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 .

Star Bright
  • 2,338
  • One can't replace that sub-expression by its limit $e$. – Paramanand Singh Mar 12 '21 at 03:02
  • @Singh Could you please elaborate why the limit theorem does not allow this? – Star Bright Mar 12 '21 at 04:09
  • 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. – Paramanand Singh Mar 12 '21 at 04:43
  • @ParamanandSingh: The first limit can be replaced by $e$ by the following: $\lim_{t\to 0} (1+t)^{\frac 1t}=e$, so for every $\epsilon\gt 0,\exists \delta_\epsilon\gt 0: |t|\lt \delta_\epsilon \implies |(1+t)^{\frac 1t} -e|\lt \epsilon$. Substituting $\Delta x/x$ for $t$ to get $|\Delta x|\lt |x|\delta_\epsilon=\delta'$ whence it follows that $|(1+\Delta x/x)^{\frac {x}{\Delta x}} -e|\lt \epsilon \implies \lim_{\Delta x\to 0} (1+\Delta x/x)^{\frac {x}{\Delta x}} =e$. And by continuity of $\log(1+s)$ at $s=e$, it follows that $\lim_{\Delta x\to 0}\log (1+\Delta x/x)^\frac {x}{\Delta x}=1$. – Koro Mar 12 '21 at 05:35
  • @Paramanad Singh Updated the justification for the change of order of limit and logarithm. – Star Bright Mar 12 '21 at 06:09
  • 1
    @koro thanks for proving the change of limit variable, just taking it for granted in my answer -:). – Star Bright Mar 12 '21 at 06:10
  • @koro: you should see the discussion I had with another answerer here in comments. The way these things work needs to be understood carefully and details should not be swept under the rug. In general an expression (which is occurring as a small part in a bigger expression whose limit needs to be evaluated) can be replaced by its limit only if occurs as a factor or as a term in overall expression. – Paramanand Singh Mar 12 '21 at 09:15
  • @Koro: also in case the expression occurs as a factor the replacement by its limit can be done only when the limit is non-zero. – Paramanand Singh Mar 12 '21 at 09:22
  • @ParamanandSingh: My only submission here is that particularly in this case $e$ can replace the first limit in this answer's 3rd last line for which I have also supplied the proof in my earlier comment in this answer. I agree with your comment in other answer wherein you rightly pointed out that $\ln(1+x)\nsim x$ as one should be very careful while making approximations and then you give a nice example with limit -1/2 which one could have mistaken to be $0$ had they used the approximation. But here $e$ can replace the first limit for a very different reason than approximation. – Koro Mar 12 '21 at 09:24
  • 1
    @Koro: I have discussed this topic at length in this answer and a formal discussion with proofs is available here. – Paramanand Singh Mar 12 '21 at 09:30
0

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}$

Koro
  • 11,402
  • 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. – Paramanand Singh Mar 12 '21 at 04:45
  • @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. – Koro Mar 12 '21 at 04:46
  • 1
    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, $o(f(x)) /f(x) \to 0$. – Paramanand Singh Mar 12 '21 at 04:51
  • @ParamanandSingh: I think $o(f(x))/f(x)\to 0$ as $f(x)\to 0$. Anyways, I'll have a look into it. Thanks. – Koro Mar 12 '21 at 04:54
  • 1
    That's not the case always. For example if $n\to\infty $ we can write $\sqrt {n} =o(n) $. – Paramanand Singh Mar 12 '21 at 04:57
  • @ParamanandSingh: Thanks. I'll learn more about little $-o$. – Koro Mar 12 '21 at 04:58