Stuck in the proof of $(1+\frac{x}{n})^n \to e^x$

Question

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

Answer 1

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.

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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 $t=0$.

Answer 2

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.

Answer 3

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

Answer 4

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 $.

Answer 5

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.