Please check my proof (L'hôpital's rule)

QuarkCharmer
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I should have made the title say "using L'h..." rather than what it is, I apologize.

Homework Statement


Prove the compounding interest formula from the following:
A=A_{0}(a+\frac{r}{n})^{nt}

Homework Equations



The Attempt at a Solution



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

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

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

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

So now the desired function is A = A_{0}e^{tu}

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

Using L'hôpital's rule gives that:

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

u = lim_{n\to\infty}\frac{rn^{2}}{n^{2}+rn}

u = lim_{n\to\infty}\frac{\frac{rn^{2}}{n^{2}}}{\frac{n^{2}}{n^{2}}+\frac{rn}{n^{2}}}

u = lim_{n\to\infty} \frac{r}{1+\frac{r}{n}} = \frac{r}{1} = r

u = r

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

A = A_{0}e^{rt}Which is the typical Pe^{rt} 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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Some remarks:

QuarkCharmer said:
u = lim_{n\to\infty} \frac{ln(1+\frac{r}{n})}{\frac{1}{n}}

Why did you drop the a?? You have exchanged a by 1. Eventually it won't matter, but I don't see how you can drop it just like that.

Using L'hôpital's rule gives that:

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

u = lim_{n\to\infty}\frac{rn^{2}}{n^{2}+rn}

This is correct, but I don't see where you get it from.

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.

Yes, you can do it, and it's one of the great tricks in working with limits. Did you discover that trick all by yourself?? Great!
 
QuarkCharmer said:
I should have made the title say "using L'h..." rather than what it is, I apologize.

Homework Statement


Prove the compounding interest formula from the following:
A=A_{0}(a+\frac{r}{n})^{nt}

Homework Equations



The Attempt at a Solution



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

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

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

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

So now the desired function is A = A_{0}e^{tu}

u = lim_{n\to\infty} \frac{ln(1+\frac{r}{n})}{\frac{1}{n}}
Why did you set a=1?
 
Oh wow, yeah that "a" should have been a "1" from the start. I typed out the formula I was starting with incorrectly and then copy/paste the errors further into the problem until I finally typed a line by hand and fixed it!

I'm a horrible latex-er?

This is correct, but I don't see where you get it from.
Truth be told, I didn't want to type out the intermediate steps where I took the derivative and simplified. I know that step to be correct and this is not a real "proof" that I am turning into anyone. I am just doing some extra problems from Stuart 6E 7.8 (This one is #85).

Aside from the missing intermediate steps, and the horrid typo (I swear it's correct on my handwritten paper), everything else is okay? More specifically, if I were to document all the steps in a proof based writing (which I have no idea how to do), it would make sense how I came to the conclusion?

Thanks for all the input.
 
QuarkCharmer said:
Aside from the missing intermediate steps, and the horrid typo (I swear it's correct on my handwritten paper), everything else is okay? More specifically, if I were to document all the steps in a proof based writing (which I have no idea how to do), it would make sense how I came to the conclusion?

Yes, everything is clear and correct!
 
micromass said:
Yes, you can do it, and it's one of the great tricks in working with limits. Did you discover that trick all by yourself?? Great!

I thought so but I doubt it. I think I remember a similar tactic used in a proof of something on khanacademy and I just stole the idea.

Thanks for the help!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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