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I found this proof for $\lim_{n \to \infty} (1+\frac{x}{n})^n=e^x$ online and wish to ask one small tiny bit of it

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

$$\lim_{n \to +\infty} \left(1 + \frac{x}{n}\right)^n =\lim_{n \to +\infty} e^{n \ln(1+\frac{x}{n})} \\ =e^{\lim_{n \to +\infty} n \ln(1+\frac{x}{n})} =e^{\lim_{n \to +\infty}\frac{ \ln(1+\frac{x}{n})}{\frac{1}{n}}}$$

Apply L'Hopital's Rule:

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

What I don't understand is the $\frac{-x}{n^2}$. Where does it come from? Isn't the derivative of $\ln(1+\frac{x}{n})$ is $\dfrac{1}{1+\frac{x}{n}}$ only?

Ant
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James Warthington
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  • You might want to plug $x = 0, n = 2$ into your formula to see that it's complete nonsense, regardless of any purported "proof" of it. – John Hughes Oct 12 '19 at 19:40
  • @JohnHughes Can you be more specific about your criticism of this so-called proof. I have tried search for one proof that is accessible to me and this one is the only one that I can comprehend. Other proofs are more involved than this one. – James Warthington Oct 12 '19 at 21:00
  • It is in here: https://math.stackexchange.com/questions/882741/limit-of-1-x-nn-when-n-tends-to-infinity/882763#882763 evinda'sanswer – James Warthington Oct 12 '19 at 21:03
  • I'm not criticizing the proof -- I'm criticizing the statement you're trying to prove, which, as stated, is simply false, as my example shows. @MohammedRiaziKermani's answer states the correct result; others have pointed to correct proofs (of the correct result, of course). – John Hughes Oct 12 '19 at 21:21

3 Answers3

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First of all your answer is $e^x$ not $e^n$ so please edit your question.

Secondly derivative of $\ln u $ is $\frac {u'}{u}$ not $1/u$ due to the chain rule of differentiation

Mohammad Riazi-Kermani
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In the limit you are differentiating w.r.t to $n$ not with $x$: $$\frac{\mathrm d}{\mathrm dn}\left[\ln\left(1+\frac{x}{n}\right)\right] = \frac{-x}{n^2}.\frac{1}{(1+\frac{x}{n})}$$

Jean-Claude Arbaut
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annoying_noob
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I have finally found another proof for this:

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

Now let $h=\frac{x}{n}$, we have $\lim_{h \to 0} e^{x\frac{\ln(1+h)}{h}}$. Since the limit $\frac{\ln(1+h)}{h}=1$, we have arrived at the final result $e^{x}$

Is this way correct?

James Warthington
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