I should have made the title say "using L'h...." rather than what it is, I apologize.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

Prove the compounding interest formula from the following:

[tex]A=A_{0}(a+\frac{r}{n})^{nt}[/tex]

2. Relevant equations

3. The attempt at a solution

[tex]A=A_{0}(a+\frac{r}{n})^{nt}[/tex]

[tex]lim_{n\to\infty}A_{0}(a+\frac{r}{n})^{nt}[/tex]

[tex]A_{0}e^{t lim_{n\to\infty}n ln(a+\frac{r}{n})}[/tex]

let u = [itex]lim_{n\to\infty}n ln(a+\frac{r}{n})[/itex]

So now the desired function is [itex]A = A_{0}e^{tu}[/itex]

[tex]u = lim_{n\to\infty} \frac{ln(1+\frac{r}{n})}{\frac{1}{n}}[/tex]

Using L'hôpital's rule gives that:

[tex]u = lim_{n\to\infty} \frac{\frac{d}{dn}ln(1+\frac{r}{n})}{\frac{d}{dn} \frac{1}{n}}[/tex]

[tex]u = lim_{n\to\infty}\frac{rn^{2}}{n^{2}+rn}[/tex]

[tex]u = lim_{n\to\infty}\frac{\frac{rn^{2}}{n^{2}}}{\frac{n^{2}}{n^{2}}+\frac{rn}{n^{2}}}[/tex]

[tex]u = lim_{n\to\infty} \frac{r}{1+\frac{r}{n}} = \frac{r}{1} = r[/tex]

[tex]u = r[/tex]

Now, re-substituting r back in for u yields:

[tex]A = A_{0}e^{rt}[/tex]

Which is the typical [itex]Pe^{rt}[/itex] formula we all know and love?

I don't want to look it up and spoil it because I think I am right. Not sure if you can raise e to the ln of something, but I don't see why not. It's not like it changes anything and I figure you have to get an e in there somehow.

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# Homework Help: Please check my proof (L'hôpital's rule)

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