Proof of the value of e without using L'Hopital's rule?

In summary, there are proofs of the limit of the compound interest function that do not involve using L'Hopital's rule. One such proof involves expanding the function by binomial theorem and using the MVT of integral calculus. This proof shows that the limit of (1 + 1/n)^n is equal to e, without having to take the derivative of the natural log function.
  • #1
bitrex
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The proofs of [tex]y = \displaystyle\lim_{x \to\infty} (1 + 1/x)^{x} = e[/tex] that I have seen basically involve taking the natural log of both sides and getting the equation in a form where L'Hopital's rule can be applied. The problem I have with this is that it requires taking the derivative of the natural log function, and the proof of the derivative of that requires prior knowledge that the limit above converges to e! It all seems a bit circular - is there a proof of the limit of the compound interest function that doesn't involve using L'Hopital's rule?
 
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  • #2
Hello,
If you expand it by binomial theorem,

[tex](1+\frac{1}{n})^n=1^n+\begin{pmatrix}n\\ 1\end{pmatrix}1^n\frac{1}{n}+\begin{pmatrix}n\\ 2\end{pmatrix}1^{n-1}(\frac{1}{n})^2+..+\begin{pmatrix}n\\ n\end{pmatrix}(\frac{1}{n})^n[/tex]
Now we look k. term

[tex]\begin{pmatrix}n\\ k\end{pmatrix}\frac{1}{n^k}=\frac{1}{k!}\frac{n(n-1)(n-2)..(n-k+1)}{n^k}\Rightarrow \lim_{n\rightarrow \infty}\begin{pmatrix}n\\ k\end{pmatrix}\frac{1}{n^k}=\frac{1}{k!}\lim_{n\rightarrow \infty}\frac{n(n-1)(n-2)..(n-k+1)}{n^k}=\frac{1}{k!}[/tex]

So our first serie equals to [tex]1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...=\lim_{n\rightarrow \infty}(1+\frac{1}{n})^n=e[/tex]
 
  • #3
[tex] log(1 + 1/n)^n = nlog(1 + 1/n) = n \int_1^{1 + 1/n} \frac{du}{u} = n*(1/n)*c [/tex] where 1 < c < 1 + 1/n. Note that I used the MVT of integral calculus.

[tex] n*(1/n)*c = c [/tex] which tends to 1 as n goes to infinity since 1 < c < 1 + 1/n. Thus [tex] \lim_{n \rightarrow \infty} log(1 + 1/n)^n = 1 [/tex] so by the continuity of the exponential function, [tex] \lim_{n \rightarrow \infty} e^{log(1+1/n)^n} = \lim_{n \rightarrow \infty} (1 + 1/n)^n = e^1 = e [/tex]
 
  • #4
Thanks, everyone!
 

FAQ: Proof of the value of e without using L'Hopital's rule?

What is the value of e?

The value of e is approximately 2.71828.

How is e defined?

e is defined as the base of the natural logarithm, where ln(e) = 1.

What is the significance of e in mathematics?

e is a fundamental constant in mathematics that appears in many important mathematical equations and concepts, such as compound interest, growth and decay, and the normal distribution.

Can you prove the value of e without using L'Hopital's rule?

Yes, there are multiple methods to prove the value of e without using L'Hopital's rule, such as using the Taylor series expansion of ex, the limit definition of e, or the Euler's formula for ex.

Why is it important to be able to prove the value of e without using L'Hopital's rule?

L'Hopital's rule is a specific method for finding limits and should not be relied upon as the only way to prove a value. Being able to prove the value of e without using this rule showcases a deeper understanding of mathematical concepts and allows for a more diverse set of problem-solving skills.

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