limits of logarithm

Question

I am trying to understand the definition of a logarithm, because when I was trying to find the derivative of $2^x$ I got $$2^x \lim_{h \to 0} \frac{2^h-1}{h}$$ which I have found by searching to be $\ln(2)$. I did get a bit confused because I would need to use l'hopital rule, which would bring be back to what I was trying to find.

But my question that I think I need to understand before getting to my second question.

Euler defines logarithm as $$\ln(x)=\lim_{n \to \infty}n(x^{\tfrac{1}{n}}-1)$$

Which then must be equal to $$\ln(x)=\lim_{h \to 0} \frac{x^h-1}{h} $$

Could you help me understand how these are both the same?

I'm not sure how Euler defined logarithm, but $\lim \limits_{n \rightarrow \infty} n(x^{\frac{1}{2}} - 1)$ is clearly not right as it is $\pm \infty$ for $x \neq 1$. It was probably $\lim \limits_{n \rightarrow \infty} n(x^{\frac{1}{n}} - 1)$. — Karolis Juodelė, Sep 09 '12 at 11:43

score 2 · Accepted Answer · answered Sep 09 '12 at 12:00

2

These are both the same since if you define $h = 1/n$, you get your second equation from the first.

answered Sep 09 '12 at 12:00

Nathaniel Bubis

33,018

Monkey D. Luffy · Answer 2 · 2012-09-09T12:19:38.930

1

If you are trying to find the derivative of $2^x$ then you also can do it using the property of $e$

$$ \lim_{h \to 0} \frac{2^h-1}{h} = \lim_{h \to 0} \frac{e^{\ln(2) h}-1}{h \ln (2)}\times \ln(2) = \ln (2) \times 1 = \ln(2)$$

edited Sep 09 '12 at 12:19

answered Sep 09 '12 at 12:01

Monkey D. Luffy

1,632

1

The problem here is that the limit appeared trying to find the derivative of $2^x$, which means using l'Hopital would get you into a circular argument. Trivia: You get into the same pickle when differentiating $\sin (x)$. – Arthur Sep 09 '12 at 12:03
@Arthur $$\frac{d}{dx}sin(x) = \lim_{h\to\infty} \frac{sin(x+h)-sin(x)}{h}$$ $$=\lim_{h\to\infty} \frac{sin(x)cos(h)+cos(x)(sin(h)-sin(x)}{h}$$ $$\because sin(x+h)=sin(x)cos(h)+cos(x)sin(h)$$ $$ =\lim_{h\to\infty}\frac{sin(x)(cos(h)-1)+cos(x)sin(h)}{h}$$ $$=\lim_{h\to\infty} sin(x)\frac{cos(h)-1}{h} + \lim_{h\to\infty} cos(x)\frac{sin(h)}{h} $$ $$ \frac{d}{dx}sin(x)=sin(x)(0) + cos(x)(1) = cos(x)$$ – yiyi Sep 09 '12 at 14:05
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@MaoYiyi First off, I think you mean $h\to 0$. Then I would like to point your attention towards the limit $$\lim_{h\to0} \cos(x)\frac{\sin(h)}{h}$$ in which one might use l'Hopital since that's easiest, without considering the circular argument. – Arthur Sep 09 '12 at 15:49
@Arthur thanks alot – yiyi Sep 09 '12 at 16:25

limits of logarithm

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