How to show that $\lim_{n\rightarrow \infty }(1+\frac{x}{n})^{n}=e^{x}$ [duplicate]

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I have a problem:

$E(x)=e^x$, $L(x)=\ln (x)$, $E^{-1}(x)=L(x)$.

Show that $\lim_{n\rightarrow \infty }(1+\frac{x}{n})^{n}=e^{x}$

Hint: use $f(t)=\ln (1+xt)$ and look at $f'(0), x\neq 0$.

I found out that $f'(0)=x$, so I can put it in $e^{f'(0)}$, but I don't know how it can help me. Please, can anyone here help me with this problem or give me a link to a good proof. I'm desperate.

marked as duplicate by AlexR, Davide Giraudo, Joel Reyes Noche, Namaste, Robert CardonaMar 10 '15 at 14:31

• You should fix $\lim_{n\to\infty}$ and not $\lim_{x\to\infty}$. – AlexR Mar 10 '15 at 11:52
• Yes, it was helpful, but I hope that there is somebody here that knows if I can use f(t) in a way that links all the functions given in a good way. But thx :) – netwon1227 Mar 10 '15 at 11:56
• I remember that when I started study maths at uni in my first year course of Calculus (or Analysis) that was our definition of exponential function. Then it was a real mess to prove that $e^{x+y}=e^xe^y$. If you are interested the only book that I know that uses this definition is by Miranda (and it is in Italian, sorry!). – rafforaffo Mar 10 '15 at 11:58
• @netwon1227 See my answer for that. – AlexR Mar 10 '15 at 12:00

you already have $f'(0) = x$ for $f(x) = \ln(1+xt)$ if we write the definition of $$x = f'(0) = \lim_{n \to \infty} \frac{f(1/n) - f(0)}{1/n} = \lim_{n \to \infty}n\left(\ln(1 + x/n)\right) = \lim_{n \to \infty}\ln((1+x/n)^n$$
exponentiating the equality gives you $$e^x = \lim_{n \to \infty}\left(1+\frac xn\right)^n.$$
• @netwon1227, the definition $f'(0) = \lim_{h \to 0}\frac{f(h) - f(0)}{h}.$ now choose $h = \frac 1n$ and let $n \to \infty$ – abel Mar 10 '15 at 13:28
Use that $\ln$ is continuous, so show that $$\lim_{n\to\infty} \ln((1+\frac xn)^n) = x$$ Where $$\ln((1+\frac xn)^n) = n\ln(1+\frac xn) = \frac{f(\frac1n)-f(0)}{\frac1n} \to f'(0) = x$$ By the definition of the derivative.