Proof of this limit formula for e

[Arnoldjavs3]

Homework Statement

http://prntscr.com/dcfe0u

Homework Equations

The Attempt at a Solution

So I'm not really strong in proofs but I think you may be able to do something like this:$$lnL = \frac{ln(1+1/x)}{x}$$
$$lnL = \frac{1/x^2}{1+1/x}$$
and then more simplifying I get something like:

$$lnL = \frac{x}{x^2+x^3}$$

At this point I think I'm just incredibly off. If I were to continue that I think I would have to multiply the right side by f(x) to remove ln from the Limit.

How do I approach this proof?

 

[fresh_42]

I can't quite follow your equations. I don't think they are correct. But to start with, what is $e$? And how is the limit defined?

 

[Stephen Tashi]

The proof will depend on how your text materials define $e$.

Edit your post and include the definition of $e$ in the "relevant equations" section.

 

[PeroK]

What definition of $e$ are you using?

 

[Mark44]

Homework Statement

http://prntscr.com/dcfe0u

Homework Equations

The Attempt at a Solution

So I'm not really strong in proofs but I think you may be able to do something like this:$$lnL = \frac{ln(1+1/x)}{x}$$

The above is incorrect. Starting with
$L = (1 + \frac 1 x)^x$
Taking the natural log of both sides yields this:
$\ln L = x \ln (1 + \frac 1 x)$
That's different from what you have above.

$$lnL = \frac{1/x^2}{1+1/x}$$
and then more simplifying I get something like:

$$lnL = \frac{x}{x^2+x^3}$$

At this point I think I'm just incredibly off. If I were to continue that I think I would have to multiply the right side by f(x) to remove ln from the Limit.

How do I approach this proof?

 

[Arnoldjavs3]

After a bit of googling I came up with a solution I think it's right(very different from the jibberish i had in OP) but I'm not sure.

I did the solution on mywhiteboard as this seems very tedious to do on mathjax.

http://prntscr.com/dchmps

After a second look that looks messy. Sorry about that lol

 

[Mark44]

After a bit of googling I came up with a solution I think it's right(very different from the jibberish i had in OP) but I'm not sure.

I did the solution on mywhiteboard as this seems very tedious to do on mathjax.

http://prntscr.com/dchmps

After a second look that looks messy. Sorry about that lol

Not only that, it doesn't make much sense. You have a lot of stuff in there that is either meaningless (such as $\lim_{x \to \infty}(1 + 0)^{\infty}$) or unrelated to this problem (such as the definition of the derivative you have in the middle).

If you continue from what I have in post #5, you should be able to finish it off. The key idea is that $\lim \ln f(x) = \ln \lim f(x)$, as long as f is continuous. IOW, you can switch the order of the limit and log operations, under suitable conditions.

 

[Ray Vickson]

After a bit of googling I came up with a solution I think it's right(very different from the jibberish i had in OP) but I'm not sure.

I did the solution on mywhiteboard as this seems very tedious to do on mathjax.

http://prntscr.com/dchmps

After a second look that looks messy. Sorry about that lol

Please type up your work; many of us will not open attachments or look at picture files. (Please read the post "Guidelines ford students and helpers", by Vela, for more information on this issue.)

 

[lurflurf]

$$\lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right)^x%=\exp\left(\lim_{x\rightarrow\infty}\frac{\log\left(1+\frac{1}{x}\right)}{\frac{1}{x}}\right)$$
use
continuity of exponential
$$\lim_{x\rightarrow\infty}e^{f(x)}=\exp\left(\lim_{x\rightarrow\infty}f(x)\right)$$
exponential identiy
$$x^y=\exp(y \log(x))$$
value of e
$$e=\exp(1)$$
and derivative of log
$$\lim_{x\rightarrow\infty}\frac{\log\left(1+\frac{1}{x}\right)}{\frac{1}{x}}=\log^\prime(1)=1$$

 

[stevmg]

The above "solutions" are all tautologies. They all assume that e exists and is the base of natural logarithms. Now, this is true, but the proof of this is outlined at
http://planetmath.org/convergenceofthesequence11nn, I have found no other site with this proof.

Essentially done in two steps.

1) (1 + 1/n)^n is expanded (binomial expansion) to n and n + 1 terms
2) comparing term by term, the n + 1 expansion is ALWAYS greater than the n expansion
3) thus the sum of the series is monotone increasing

4) the second step is to show that for any n expansion, even as n ⇒ ∞, the sum is always less than 3. Hence the sum of the series is bounded. Monotone increasing with an upper bound ⇒ a limit.

The calculation for e would be either a Taylor series or the use of an Excel spreadsheet. With the latter, one would, through iteration, obtain results until the results do not change, which would be the limit of the expansion that Excel could produce.

That is a more rigorous proof and does not require one to use a tautology to come up with an answer.