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How do you prove ${\lim\limits_{m \to +\infty}} \ln{\left(1+\frac{r}{m}\right)} = \frac{r}{m}$, where $0 < r < 1$?

I encountered this problem while trying to reproduce the third to last step in this proof.

ViktorStein
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megamonium
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    How can the changing variable in the limit ($m$) appear outside the limit? – Parcly Taxel Jul 14 '18 at 09:06
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    See https://math.stackexchange.com/questions/690898/determine-the-following-limit-as-x-approaches-0-frac-ln1xx – Robert Z Jul 14 '18 at 09:06
  • What are the ideas / approaches you have tried? – ViktorStein Jul 14 '18 at 09:07
  • I looked at the link you gave and it says: ''as $m$ gets really large, $\frac{r}{m}$ gets really small, so we can use the log approximation $\ln(1 + h) \approx h$.''

    Notice that you can only evaluate a limit of a sum as the sum of the limits if you know the limits exist.

     – ViktorStein Jul 14 '18 at 09:13
  • I'm sorry, but this question is not the same as the one referred to. It is linked, but here there is clearly a misunderstanding of the notion of limit. So I vote to reopen it. – Bernard Jul 14 '18 at 09:17

3 Answers3

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You don't. As $m$ is inside a limit on the left side of your equality, there should be no $m$ on the right side.

What you can say is that $\lim_{m\to+\infty}\frac{\ln(1+\frac rm)}{\frac rm} = 1$, and you can prove it by studying function $f;x\mapsto \frac{\ln (1+x)-x}{x}$.

Nicolas FRANCOIS
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$$\lim_{m\to+\infty} \ln\left(1+\frac{r}{m}\right) = \ln(1) = 0$$

$m$ goes to infinity, so the resulting limit can't have $m$ in it.

What's used in the proof is that $\ln(1+h) \sim h$, which is the same as saying $\lim_{h\to 0}\frac{\ln(1+h)}{h} = 1$

F.A.
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This formula is meaningless: a limit cannot depend on the dummy variable.

What is meant is probably that both sides are equivalent, denoted $$ \ln\Bigl(1+\dfrac{r}{m}\Bigr) \sim_{m\to\infty} \frac{r}{m}$$ in the sense that $$\lim_{m\to\infty}\smash[t]{\frac{\ln\Bigl(1+\dfrac{r}{m}\Bigr)}{\dfrac{r}{m}}=1}$$ (which does not imply the difference tends to $0$).

Bernard
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