Compound interest ($r$ in terms of $\frac{dA}{dt}$)

Question

Suppose the amount of money in bank account that is compounded annually is given by $A(t)$. The annual rate of interest is $r$. Find a relation between $\displaystyle\frac{dA}{dt}$ and $r$.

My attempt:

$A(t)=A(0)(1+r)^t$

$\displaystyle\frac{dA}{dt}=A(0)(1+r)^t\ln (1+r)$

Dividing, we get

$\displaystyle\frac{1}{A}\frac{dA}{dt}=\ln(1+r)$

However, I think the answer should be $\displaystyle r=\frac{1}{A}\frac{dA}{dt}$.

What am I doing wrong?

Answer 1

You are not doing anything wrong. Since $$ \lim_{r \to 0} \frac{\ln(1+r)}{r} = 1 $$ (to see why: use l'Hôpital's rule or Taylor approximation) holds, for small numbers $r$ $$ \ln(1+r) \approx r. $$