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

In summary, the compounding interest formula A=A_{0}(a+\frac{r}{n})^{nt} can be proven by taking the limit as n approaches infinity of A=A_{0}(a+\frac{r}{n})^{nt}, substituting u for the limit of n ln(a+\frac{r}{n}), using L'Hôpital's rule to simplify, and re-substituting the value of u to get A=A_{0}e^{rt}. This is a useful trick for working with limits and was discovered through the help of resources such as Khan Academy.
  • #1
QuarkCharmer
1,051
3
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:
[tex]A=A_{0}(a+\frac{r}{n})^{nt}[/tex]

Homework Equations



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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  • #2
Some remarks:

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

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:

[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]

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!
 
  • #3
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:
[tex]A=A_{0}(a+\frac{r}{n})^{nt}[/tex]

Homework Equations



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]
Why did you set a=1?
 
  • #4
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.
 
  • #5
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!
 
  • #6
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!
 

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

What is L'hôpital's rule?

L'hôpital's rule is a mathematical theorem used to evaluate limits involving indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of the ratio of two functions is of an indeterminate form, then the limit of the ratio of their derivatives will be the same.

When should L'hôpital's rule be used?

L'hôpital's rule should be used when evaluating limits that result in an indeterminate form. This typically occurs when the limit involves a fraction with a denominator that approaches 0 and a numerator that approaches either 0 or infinity.

What are the steps for using L'hôpital's rule?

The steps for using L'hôpital's rule are as follows:

  1. Determine the limit of the ratio of the two functions.
  2. If the result is an indeterminate form, take the derivative of both the numerator and denominator.
  3. Evaluate the new limit using the derivatives.
  4. If the result is still an indeterminate form, repeat the process until the limit is no longer indeterminate.

Are there any restrictions for using L'hôpital's rule?

Yes, there are a few restrictions for using L'hôpital's rule. The functions involved must be differentiable and the limit must be in the form of 0/0 or ∞/∞. Additionally, the limit must approach either 0 or infinity.

Can L'hôpital's rule be used for other types of limits?

No, L'hôpital's rule is specifically designed for limits involving indeterminate forms. It cannot be used for other types of limits, such as limits at infinity or limits involving trigonometric functions.

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