Is it possible to prove it using derivative first principle?

I know how to prove lim n->max+ [ 1 + 1/n ]^n = e

but I am not sure how to prove the captioned one.

thank you very much.

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- March 19th 2014, 02:59 PMasdqweGiven lim h -> 0, how to prove: ln(1 + hr) / h = r ?
Is it possible to prove it using derivative first principle?

I know how to prove lim n->max+ [ 1 + 1/n ]^n = e

but I am not sure how to prove the captioned one.

thank you very much. - March 19th 2014, 06:19 PMjohngRe: Given lim h -> 0, how to prove: ln(1 + hr) / h = r ?
Hi,

To show:

$$\lim\limits_{h\to0}{\ln(1+hr)\over h}=r$$

First suppose r = 0. Then the numerator is identically 0 and so the limit is r = 0. Otherwise, first note the derivative of $\ln(x)$ at $x=1$ is 1. That is,

$$1=\lim\limits_{k\to0}{\ln(1+k)-ln(1)\over k}=\lim\limits_{k\to0}{\ln(1+k)\over k}$$

Now let $k=hr$. Then

$$\lim\limits_{h\to0}{\ln(1+hr)\over h}=r\cdot\lim\limits_{h\to0}{\ln(1+hr)\over hr}=r\cdot\lim\limits_{k\to0}{\ln(1+k)\over k}=r\cdot1=r$$

If the switch from "h approaches 0" to "k approaches 0" is mysterious, informally, when h approaches 0, k approaches 0. More formally, an $\epsilon$ argument can be given. If you have question, I'll try to answer. - March 19th 2014, 08:33 PMSorobanRe: Given lim h -> 0, how to prove: ln(1 + hr) / h = r ?
Hello, asdqwe!

Quote:

Is it possible to prove it using derivative first principle?

You have to give us a function to differentiate first.

I know how to prove: .

But I am not sure how to prove: .

Multiply by

Let

Then we have: .

. . . . . . . . . . . - March 20th 2014, 04:05 AMasdqweRe: Given lim h -> 0, how to prove: ln(1 + hr) / h = r ?
Thank you for helping me. Learned a lot.