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Does anyone know the exact proof of this limit result?

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


marked as duplicate by Thomas Andrews, Namaste, Daniel Fischer, user940, M Turgeon Jul 30 '14 at 16:25

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    $\begingroup$ Depends on the definition of $e^x$... $\endgroup$ – Thomas Andrews Jul 30 '14 at 15:32
  • $\begingroup$ have you tried to explicit the power and take the limit? It converge to the taylor series. $\endgroup$ – Lolman Jul 30 '14 at 15:33
  • $\begingroup$ I tried by taking log of both sides, but I don't know what to do after this step. Thought of using L'Hopital's rule. But that ain't helping. $\endgroup$ – narendra-choudhary Jul 30 '14 at 15:38
  • $\begingroup$ You can use this technique. $\endgroup$ – Mhenni Benghorbal Jul 30 '14 at 15:44
  • 4
    $\begingroup$ Using L'Hopital on the log you get: $$\lim_{n\to\infty} \frac{\log(1+\frac{x}{n})}{1/n}=\lim_{n\to\infty}\frac{n}{n+x}\frac{-x}{n^2}(- n^2)=x$$ $\endgroup$ – Lolman Jul 30 '14 at 15:46

A short proof:

$\left(1+\frac{x}{n}\right)^n = e^{n\log\left(1+\dfrac{x}{n}\right)}$

Since $\log(1+x) = x + O(x^2)$ when $x \to 0$, we have $n\log(1 + \frac{x}{n}) = x + O(\frac{x^2}{n})$ when $n\to +\infty$

  • $\begingroup$ neat, really good $\endgroup$ – Lolman Jul 30 '14 at 15:37
  • 3
    $\begingroup$ Or just use the derivate of $\log$ to evaluate $\lim_{y\to 0}\frac{\log(1+xy)}{y}$, then let $y=1/n$ to get the limit of $n\log(1+x/n)$... $\endgroup$ – Thomas Andrews Jul 30 '14 at 15:39
  • $\begingroup$ Even more elementary would be to use the basic inequality $\frac{x/n}{1+x/n}\le \log(1+x/n)\le x/n$. Then a straightforward application of the squeeze theorem does the trick. $\endgroup$ – Mark Viola Oct 25 '15 at 16:03
  • $\begingroup$ In this case, why does writing $(1+\frac{x}{n})^n=10^{n\log_{\, 10} \, \, (1+\frac{x}{n})}$ not show that $\lim_{n \to \infty}(1+\frac{x}{n})^n=10^x$? $\endgroup$ – Rasputin Jan 19 at 21:17
  • $\begingroup$ @Rasputin because the Taylor expansion of $\log_{10} (1+x)$ is different $\endgroup$ – Petite Etincelle Jan 20 at 13:27

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

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

Apply L'Hopital's Rule:

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

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

  • $\begingroup$ First: Is the first result in the fourth row by L'Hopital's rule? Second, isn't a limit missing in the second result of the fourth row? $exp(x/(1+x/n))$ should be $exp\{lim[x/(1+x/n)]\}$ $\endgroup$ – Carol Eisen Dec 25 '17 at 17:39
  • $\begingroup$ You are correct! I edited it to account for the missing limit. Also, you are correct in that one of the steps is using L'Hopital's Rule. $\endgroup$ – H. Dewey May 2 '18 at 19:58
  • $\begingroup$ Should I have permission to apply L' Hospital as it is n tends to infinity not x ,I mean it is kind of discrite case not continuous. $\endgroup$ – Supriyo Banerjee Apr 24 at 14:53

You can use the binomial series expansion. For example:

$$\left(1+\frac{x}{n}\right)^n =1+ \frac{n}{1!}\left(\frac{x}{n}\right)^1+\frac{n(n-1)}{2!}\left(\frac{x}{n}\right)^2+\frac{n(n-1)(n-2)}{3!}\left(\frac{x}{n}\right)^3+\cdots $$

$$\left(1+\frac{x}{n}\right)^n =1+ \frac{n}{n}x+\frac{n(n-1)}{n^2}\frac{x^2}{2!}+\frac{n(n-1)(n-2)}{n^3}\frac{x^3}{3!} + \cdots$$

As $n \to \infty$ the coefficients in $n$ all tend to $1$. Hence:

$$\lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots $$ You'll recognise this last power series as the Taylor series for $\mathrm{e}^x$.

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    $\begingroup$ what if there are infinite terms and then you cannot exchange the sum and limit if n is non integer $\endgroup$ – happymath Mar 3 '15 at 10:17
  • $\begingroup$ it's not easy to make this argument rigorous. This answer would be better if it pointed out that fact $\endgroup$ – jkabrg May 30 at 18:28
  • $\begingroup$ a rigorous version of this argument uses the monotone convergence theorem $\endgroup$ – jkabrg May 30 at 18:41

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