# Continuous compounding problem [duplicate]

How long does it take an investment to double if continuously compounded at a 6% rate? Is this right?

\begin{align*} e^{0.06 \cdot t} & = 2\\ 0.06t & = \ln(2)\\ t & = \frac{\ln 2}{0.06}\\ & \approx 11.55 \end{align*}

Why does $e$ equal the $\lim_{n \to \infty} (1+\frac{1}{m})^m$?

I know that if $r$ is the rate and $n$ is the number of compounding periods in $t$ years, then the value of an investment is:

$$A_0\left(1+\frac{r}{n}\right)^{nt}$$

But how do we get from here to the definition with $e$ in continuous compounding?

## marked as duplicate by user296602, N. F. Taussig, Misha Lavrov, Saad, ShaileshFeb 19 '18 at 0:33

• The question of why $e = \lim_{n \to \infty} (1 + 1/n)^n$ has been asked here a very large number of times. – user296602 Feb 12 '18 at 1:31
$lim_{n\rightarrow \infty}(1+\frac{r}{n})^n = e^r$ is proved in calculus. If you don't remember this, you can prove it easily by proving that the logarithm goes to $r$. As to your second question, $(1+\frac{r}{n})^{nt}=((1+\frac{r}{n})^n)^t,$ by the laws of exponents.