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.
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.
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.
Hello, asdqwe!
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: .
. . . . . . . . . . .