# Math Help - Given lim h -> 0, how to prove: ln(1 + hr) / h = r ?

1. ## Given 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.

2. ## Re: 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.

3. ## Re: Given lim h -> 0, how to prove: ln(1 + hr) / h = r ?

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: . $\lim_{n\to\infty}\left(1 + \tfrac{1}{n}\right)^n \:=\: e$

But I am not sure how to prove: . $\lim_{h\to0} \frac{\ln(1+hr)}{h} \:=\:r$

$\text{We have: }\:\lim_{h\to0}\frac{\ln(1+hr)}{h} \;=\;\lim_{h\to0} \left[\frac{1}{h}\ln(1+hr)\right]$

Multiply by $\tfrac{r}{r}\!:\;\;\lim_{h\to0}\left[\frac{r}{hr}\ln(1+hr)\right] \;=\;\lim_{h\to0}\ln(1 + hr)^{\frac{r}{hr}} \;=\;\lim_{h\to0}\ln\left[(1+hr)^{\frac{1}{hr}}\right]^r$

Let $n = \tfrac{1}{hr}$
Then we have: . $\lim_{n\to\infty}\ln\left[\left(1 + \tfrac{1}{n}\right)^n\right]^r \;=\;\ln\left[\lim_{n\to\infty}\left(1+\tfrac{1}{n}\right)^n \right]^r$

. . . . . . . . . . . $=\;\ln(e)^r \;=\;r\cdot\ln(e) \;=\;r$

4. ## Re: Given lim h -> 0, how to prove: ln(1 + hr) / h = r ?

Thank you for helping me. Learned a lot.

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