how do you differentiate $\ln(x)$ using the difference quotient.

Question

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?

No matter what the definition one thing you do get from that calculation is $\log'(x) = \frac1x \log'(1)$. — hmakholm left over Monica, Jun 29 '19 at 13:32
@yay I posted a solution that relies simple bounds for the logarithm, which can be found using elementary (i.e., pre-calculus) analysis only. — Mark Viola, Jun 29 '19 at 15:14
You need to include your definition of the logarithmic function in order for a clear non-circular answer. — Nap D. Lover, Jun 29 '19 at 17:49
my definition of the logarithmic function is http://mathworld.wolfram.com/NaturalLogarithm.html — Ekadh Singh - Reinstate Monica, Jun 29 '19 at 20:58

score 5 · Answer 1 · answered Jun 29 '19 at 13:35

5

For finite non-zero $x,$ $$\lim_{h\to0}\dfrac{\ln\left(1+\dfrac hx\right)}h=\dfrac1x\cdot\lim_{h\to0}\dfrac{\ln\left(1+\dfrac hx\right)}{\dfrac hx}=?$$

answered Jun 29 '19 at 13:35

lab bhattacharjee

274,582

And how does one evaluate that limit? That is the point of the problem. – Mark Viola Jun 29 '19 at 15:09
@MarkViola, https://math.stackexchange.com/questions/650502/evaluate-lim-limits-y-to-0-log1y-y-without-lhr-or-taylor-series – lab bhattacharjee Jun 29 '19 at 15:14
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. – Mark Viola Jun 29 '19 at 15:17

Mark Viola · Accepted Answer · 2019-07-08T21:42:27.863

5

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

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

edited Jul 08 '19 at 21:42

answered Jun 29 '19 at 15:04

Mark Viola

179,405

Am I misunderstanding something, or shouldn't the limit in the last equation go to $0$? – WilliamMorris Jul 03 '19 at 18:26
I don't understand. What equation is the "last equation?" – Mark Viola Jul 03 '19 at 18:54
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? – WilliamMorris Jul 08 '19 at 16:53
Yes. $h\to 0$ is correct. – Mark Viola Jul 08 '19 at 21:41

B. Goddard · Answer 3 · 2019-06-29T14:13:04.733

3

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

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

edited Jun 29 '19 at 14:13

answered Jun 29 '19 at 13:38

B. Goddard

1

1

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. – Mark Viola Jun 29 '19 at 15:12

score 2 · Answer 4 · answered Jun 29 '19 at 13:35

2

$$\lim_{h\to 0} \frac{\ln(1+\frac hx)}{h}=\frac1x\lim_{\frac hx \to 0}\frac{\ln(1+\frac hx)}{\frac hx}=\frac1x$$

answered Jun 29 '19 at 13:35

Amit Rajaraman

3,248

how do we go from the limit of h to 0 to the limit of h/x to 0? – Ekadh Singh - Reinstate Monica Jun 29 '19 at 13:43
If $h\to0$ and $x$ is non-zero (it is fixed), then $\frac hx\to0$. – Amit Rajaraman Jun 29 '19 at 13:45
The important fact here is that $x$ is held constant while we take the limit over $h$. – hmakholm left over Monica Jun 29 '19 at 13:46
@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? – Mnifldz Jun 29 '19 at 14:10
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 $1$. – Amit Rajaraman Jun 29 '19 at 14:18

score 1 · Answer 5 · answered Jun 29 '19 at 13:35

1

Multiply and divide the denominator by $x$. Clearly we won't differentiate at $x=0$.

So,

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

answered Jun 29 '19 at 13:35

19aksh

12,768

And how does one evaluate that limit? That is the point of the problem. – Mark Viola Jun 29 '19 at 15:12

Peter Szilas · Answer 6 · 2019-06-29T14:06:30.447

0

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 $h \rightarrow 0.$

edited Jun 29 '19 at 14:06

answered Jun 29 '19 at 13:52

Peter Szilas

20,344
2
17
28

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. – Mark Viola Jun 29 '19 at 15:13
Agreed.Thought it could complement the given answers. – Peter Szilas Jun 29 '19 at 15:28

how do you differentiate $\ln(x)$ using the difference quotient.

6 Answers6