Prove ${\lim\limits_{m \to +\infty}} \ln{\left(1+\frac{r}{m}\right)} = \frac{r}{m}$

Question

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.

Answer 1

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 $f;x\mapsto \frac{\ln (1+x)-x}{x}$.

Answer 2

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

$m$ 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 $\lim_{h\to 0}\frac{\ln(1+h)}{h} = 1$

Answer 3

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 $0$).