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This question already has an answer here:

So the limit is as $h$ approaches $0$ of $\displaystyle \frac{\ln(x+h)-\ln(x)}{h}$, which simplifies to $\displaystyle\frac{\ln\left(\frac{x+h}{x}\right)}{h}$, which simplifies to $\displaystyle\frac{\ln\left(1+ \frac hx\right)}{h}$.

I got stuck here. , how should I continue?

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marked as duplicate by Jam, Paul Frost, YuiTo Cheng, The Count, Xander Henderson Jul 1 at 0:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ How is you natural logarithm defined in the first place? $\endgroup$ – Henning Makholm Jun 29 at 13:30
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    $\begingroup$ No matter what the definition one thing you do get from that calculation is $\log'(x) = \frac1x \log'(1)$. $\endgroup$ – Henning Makholm Jun 29 at 13:32
  • $\begingroup$ @yay I posted a solution that relies simple bounds for the logarithm, which can be found using elementary (i.e., pre-calculus) analysis only. $\endgroup$ – Mark Viola Jun 29 at 15:14
  • $\begingroup$ You need to include your definition of the logarithmic function in order for a clear non-circular answer. $\endgroup$ – Nap D. Lover Jun 29 at 17:49
  • $\begingroup$ my definition of the logarithmic function is mathworld.wolfram.com/NaturalLogarithm.html $\endgroup$ – Yay Jun 29 at 20:58
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In THIS ANSWER, I used only the limit definition of the exponential function and Bernoulli's Inequality to show that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\ge \log(x)\le x-1\tag1$$

Hence, using $\log(x+h)-\log(x)=\log\left(1+\frac hx\right)$, we from $(1)$ we have

$$\frac{\frac1x}{1+\frac hx}\le \frac{\log\left(1+\frac hx\right)}{h}\le \frac1x\tag2$$

Applying the squeeze theorem to $(2)$ we find the coveted limit

$$\lim_{h\to 0}\frac{\log(x+h)-\log(x)}{h}=\frac1x$$

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  • $\begingroup$ Am I misunderstanding something, or shouldn't the limit in the last equation go to $0$? $\endgroup$ – WilliamMorris Jul 3 at 18:26
  • $\begingroup$ I don't understand. What equation is the "last equation?" $\endgroup$ – Mark Viola Jul 3 at 18:54
  • $\begingroup$ Apologies. I meant the limit "$\lim_{h\to \infty}\frac{\log(x+h)-\log(x)}{h}=\frac1x$". Shouldn't the limit of $h$ go to $0$, like in the definition of the derivative? $\endgroup$ – WilliamMorris Jul 8 at 16:53
  • $\begingroup$ Yes. $h\to 0$ is correct. $\endgroup$ – Mark Viola Jul 8 at 21:41
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For finite non-zero $x,$ $$\lim_{h\to0}\dfrac{\ln\left(1+\dfrac hx\right)}h=\dfrac1x\cdot\lim_{h\to0}\dfrac{\ln\left(1+\dfrac hx\right)}{\dfrac hx}=?$$

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  • $\begingroup$ And how does one evaluate that limit? That is the point of the problem. $\endgroup$ – Mark Viola Jun 29 at 15:09
  • $\begingroup$ @MarkViola, math.stackexchange.com/questions/650502/… $\endgroup$ – lab bhattacharjee Jun 29 at 15:14
  • $\begingroup$ And neither of the answers posted in that reference are satisfactory as they rely on circular logic. I just posted a solution on this page that relies only on simple bounds for the logarithm, which can be found using elementary (i.e., pre-calculus) analysis only. $\endgroup$ – Mark Viola Jun 29 at 15:17
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$$\lim_{h\to 0} \frac{\ln(1+\frac hx)}{h}=\frac1x\lim_{\frac hx \to 0}\frac{\ln(1+\frac hx)}{\frac hx}=\frac1x$$

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  • $\begingroup$ how do we go from the limit of h to 0 to the limit of h/x to 0? $\endgroup$ – Yay Jun 29 at 13:43
  • $\begingroup$ If $h\to0$ and $x$ is non-zero (it is fixed), then $\frac hx\to0$. $\endgroup$ – Epiksalad Jun 29 at 13:45
  • $\begingroup$ The important fact here is that $x$ is held constant while we take the limit over $h$. $\endgroup$ – Henning Makholm Jun 29 at 13:46
  • $\begingroup$ @Epiksalad your answer doesn't really show why the limit works. As written, your limit is still indeterminate because you would get $\frac{0}{0}$. Can you offer a more thorough explanation? $\endgroup$ – Mnifldz Jun 29 at 14:10
  • $\begingroup$ Since the limit as $x\to0$ of $(1+x)^{1/x}$ is e, if we take the logarithm on both sides, we get that the desired limit is equal to $1$. $\endgroup$ – Epiksalad Jun 29 at 14:18
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If the definition of $\ln x$ is

$$\ln x = \int_1^x \frac{1}{t} \; dt$$

Then you can write your limit as

$$\lim_{h \to 0 } \frac{1}{h} \int_x^{x+h} \frac{1}{t} \; dt.$$

Then note that for very small $h$, the area represented by the integral is nearly a rectangle of width $h$ and height $1/x$, so the limit equals

$$\lim_{h \to 0} \frac{1}{h}(h/x) = \frac{1}{x}.$$

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  • 1
    $\begingroup$ Of course this approach is the one I would take. But the OP might not have learned anything about integrals yet. So, I posted a solution that relies simple bounds for the logarithm, which can be found using elementary (i.e., pre-calculus) analysis only. $\endgroup$ – Mark Viola Jun 29 at 15:12
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Multiply and divide the denominator by $x$. Clearly we won't differentiate at $x=0$.

So,

$$L = \lim_{h\to0}\frac{\ln(1+\frac{h}{x})}{x\cdot\frac{h}{x}} = \frac{1}{x}\lim_{\frac{h}{x}\to0}\frac{\ln(1+\frac{h}{x})}{\frac{h}{x}} = \frac{1}{x}\cdot1 = \frac{1}{x}$$

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  • $\begingroup$ And how does one evaluate that limit? That is the point of the problem. $\endgroup$ – Mark Viola Jun 29 at 15:12
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Option:

$x >0$

$\log x=\displaystyle{\int_{1}^{x}} (1/t )dt.$

MVT for integration.

$(1/h)\dfrac{\log (x+h)-\log x}{h}=$

$(1/h)\displaystyle{\int_{x}^{x+h}}(1/t)dt=$

$(1/h)(1/s)\displaystyle{\int_{x}^{x+h}}1 dt=$

$(1/h)(1/s)h$; where $s \in [x,x+h].$

Take the limit $h \rightarrow 0.$

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  • $\begingroup$ Of course this approach is the one I would take. But the OP might not have learned anything about integrals yet. So, I posted a solution that relies simple bounds for the logarithm, which can be found using elementary (i.e., pre-calculus) analysis only. $\endgroup$ – Mark Viola Jun 29 at 15:13
  • $\begingroup$ Agreed.Thought it could complement the given answers. $\endgroup$ – Peter Szilas Jun 29 at 15:28

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