0

I toke this proof from Limit of $(1+ x/n)^n$ when $n$ tends to infinity and there is a step that I don't understand.

Why can we have the pink equality??

$$e^{\ln{(1 + \frac{x}{n})^n} }=e^{n \ln(1+\frac{x}{n})}$$

$$\lim_{n \to +\infty} (1 + \frac{x}{n})^n=\lim_{n \to +\infty} e^{n \ln(1+\frac{x}{n})} \\ =e^{\lim_{n \to +\infty} n \ln(1+\frac{x}{n})}=e^{\lim_{ \to +\infty}\frac{ \ln(1+\frac{x}{n})}{\frac{1}{n}}} \\ \color{fuchsia}=e^{\lim_{ \to +\infty}\frac{(\frac{-x}{n^2})\frac{1}{1+\frac{x}{n}}}{-\frac{1}{n^2}}}=e^{\frac{x}{1+\frac{x}{n}}}=e^x$$

Therefore, $$(1+\frac{x}{n})^n \to e^x$$

Aaron Martinez
  • 659

5 Answers5

3

If you know the generalization of the well known limit

$$\lim_{n\to\infty}\left(1+\frac1{a_n}\right)^{a_n}=e\;\;\;\text{whenever the sequence $\;\{a_n\}\;$ fulfills}\;\;\lim_{n\to\infty}a_n=\infty$$

then you can do as follows (beside l'Hospital, as shown in what you wrote):

$$\left(1+\frac xn\right)^n=\left[\left(1+\frac1{\frac nx}\right)^{n/x}\right]^x\xrightarrow[n\to\infty]{}e^x$$

using the sequence $\;a_n:=\cfrac nx\xrightarrow[n\to\infty]{}\infty\;,\;\;x>0\;$

For $\;x<0\;$ it is very similar. Try it.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
2

This is L'Hopital's rule (w.r.t. $n$): letting $f(n) = \ln(1+\frac{x}{n})$, $g(n) = \frac{1}{n}$, you have $$ \lim_{n\to\infty} \frac{\ln(1+\frac{x}{n})}{\frac{1}{n}} = \lim_{n\to\infty} \frac{f(n)}{g(n)} = \lim_{n\to\infty} \frac{f'(n)}{g'(n)} = \lim_{n\to\infty} \frac{\frac{-x}{n^2}\frac{1}{1+\frac{x}{n}}}{-\frac{1}{n^2}} $$ as stated.

If you ask me, though, this is an overcomplicated proof. Better to immediately recognize that $$ \lim_{n\to\infty} \frac{\ln(1+\frac{x}{n})}{\frac{1}{n}} = \lim_{n\to\infty} \frac{\ln(1+\frac{x}{n})-\ln(1+0)}{\frac{1}{n}-0} = \lim_{t\to0} \frac{\ln(1+tx)-\ln(1+0)}{t-0} = \frac{x}{1+x\cdot 0} $$ by spotting the derivative of $t\mapsto \ln(1+xt)$ at $t=0$.

Simply Beautiful Art
  • 74,685
Clement C.
  • 67,323
  • Don't you have to justify the use of L'Hôpital's rule in this sequence context? In general I don't think it's true that $$\lim_{n\to\infty}\frac{f(n)}{g(n)}=\lim_{n\to\infty}\frac{f'(n)}{g'(n)}$$ For example, consider $f(n):=\sin(2\pi n)$ and $g(n):=n$. – NeedForHelp Jun 04 '17 at 21:20
  • @NeedForHelp Not sure what you mean. L'Hopital's rule states that (under some mild assumptions) if the limit of $\frac{f'}{g'}$ exists, so does that of $\frac{f}{g}$, and they are equal. The "sequence" part is irrelevant here. – Clement C. Jun 04 '17 at 21:22
  • Isn't L'Hôpital's rule for limit of functions, not sequences? In the example I gave, $f'(n)=2\pi\cos(2\pi n)=2\pi$ and $g'(n)=1$ so you seem to claim that $$0=\lim_{n\to\infty}\frac{\sin(2\pi n)}{n}=\lim_{n\to\infty}\frac{2\pi}{1}=2\pi$$ – NeedForHelp Jun 04 '17 at 21:26
  • @NeedForHelp Replace $\frac{1}{n}$ by $x$ in the OP's limit and you'll see why it works. The limit exists when $x\to 0$, so in particular it exists along the particular sequence $(1/n)_n$ which also goes to 0. Regarding your counterexample, it is not an actual counterexample. The limit $f'(n)/g'(n)$ here does not exist! (the numerator does not converge to $2\pi$, $2\pi\cos (2\pi n)$ keeps oscillating when $n\to\infty$). – Clement C. Jun 04 '17 at 21:28
  • This little justification you just gave is what I thought was missing. I don't see how $\cos(2\pi n)$ keeps oscillating when $n\to\infty$ since it is constant. – NeedForHelp Jun 04 '17 at 21:31
  • @NeedForHelp Oh, my bad. I must have fried my last remaining neurons before that last comment. But basically, you cannot apply L'Hopital's rule to your counterexample (check the condition that ($f\to \pm\infty$ and $g\to \pm\infty$) (or $f\to 0$ and $g\to 0$). So, in short: the "sequentialization" of the limit is irrelevant to using L'Hopital's rule here, since you want to apply it to the continuous "version" (and if the limit exists when $x\to 0$, it exists when you take a particular sequence $x_n\to0$, here $(1/n)_n$.) – Clement C. Jun 04 '17 at 21:36
2

From $\ln(1+z) =\int_0^z \frac{dt}{1+t} $ and, for $t > 0$ we have $1-t \lt \frac1{1+t} \lt 1 $ we have $\ln(1+z) \gt \int_0^z (1-t) dt =z-\frac{z^2}{2} $ and $\ln(1+z) \lt \int_0^z (1) dt =z $.

Putting $z = \frac{x}{n}$, we get $\frac{x}{n}-\frac{x^2}{2n^2} \lt \ln(1+\frac{x}{n}) \lt\frac{x}{n} $ so that $x-\frac{x^2}{2n} \lt n\ln(1+\frac{x}{n}) \lt x $.

marty cohen
  • 107,799
1

L'Hopital's rule has been applied: http://mathworld.wolfram.com/LHospitalsRule.html

applyb
  • 326
1

Just another way to show it.

Consider $$A_n=\left(1+\frac{x}{n}\right)^n\implies \log(A_n)=n \log\left(1+\frac{x}{n}\right)$$ Now, since $n$ is large, by Taylor $$\log\left(1+\frac{x}{n}\right)=\frac{x}{n}-\frac{x^2}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(A_n)={x}-\frac{x^2}{2 n}+O\left(\frac{1}{n^2}\right)$$ Taylor again $$A_n=e^{\log(A_n)}=e^x-\frac{e^x x^2}{2 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.

Claude Leibovici
  • 260,315