Limit Of $(1+ X/n)^n$ When $n$ Tends To Infinity - Math Stack Exchange

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Learn more about Teams Limit of $(1+ x/n)^n$ when $n$ tends to infinity [duplicate] Ask Question Asked 10 years, 4 months ago Modified 6 years, 7 months ago Viewed 380k times 49 $\begingroup$ This question already has answers here: About $\lim \left(1+\frac {x}{n}\right)^n$ (14 answers) Closed 10 years ago.

Does anyone know the exact proof of this limit result?

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

• limits
• proof-writing

Share Cite Follow edited Jul 30, 2014 at 15:32 Darth Geek 12.4k11 gold badge3434 silver badges5252 bronze badges asked Jul 30, 2014 at 15:30 narendra-choudharynarendra-choudhary 78811 gold badge88 silver badges1414 bronze badges $\endgroup$ 6

• 2 $\begingroup$ Depends on the definition of $e^x$... $\endgroup$ – Thomas Andrews Commented Jul 30, 2014 at 15:32
• $\begingroup$ have you tried to explicit the power and take the limit? It converge to the taylor series. $\endgroup$ – Lolman Commented Jul 30, 2014 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 Commented Jul 30, 2014 at 15:38
• $\begingroup$ You can use this technique. $\endgroup$ – Mhenni Benghorbal Commented Jul 30, 2014 at 15:44
• 5 $\begingroup$ 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 Commented Jul 30, 2014 at 15:46

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3 Answers 3

Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 84 $\begingroup$

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 $n\to +\infty$

Share Cite Follow edited Jul 30, 2014 at 15:37 Thomas Andrews 180k1818 gold badges219219 silver badges413413 bronze badges answered Jul 30, 2014 at 15:36 Petite EtincellePetite Etincelle 14.9k22 gold badges3333 silver badges6161 bronze badges $\endgroup$ 7

• 4 $\begingroup$ 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 Commented Jul 30, 2014 at 15:39
• 2 $\begingroup$ 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 Commented Oct 25, 2015 at 16:03
• 2 $\begingroup$ 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 Commented Jan 19, 2019 at 21:17
• 6 $\begingroup$ @Rasputin because the Taylor expansion of $\log_{10} (1+x)$ is different $\endgroup$ – Petite Etincelle Commented Jan 20, 2019 at 13:27
• 1 $\begingroup$ What is $O(x^2)$? $\endgroup$ – PGupta Commented Sep 19, 2020 at 17:00

| Show 2 more comments 52 $\begingroup$

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

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

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

Share Cite Follow edited May 2, 2018 at 20:27 CommunityBot 1 answered Jul 30, 2014 at 15:46 evindaevinda 7,95966 gold badges4242 silver badges8888 bronze badges $\endgroup$ 6

• 2 $\begingroup$ 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 Commented Dec 25, 2017 at 17:39
• 1 $\begingroup$ 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 Commented May 2, 2018 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 Commented Apr 24, 2019 at 14:53
• $\begingroup$ I understand everything except the part $\dfrac{-x}{n^{2}}$. Where does this come from? Isn't the derivative of $\ln(1+\frac{x}{n})=\frac{1}{1+\frac{x}{n}}$ only? $\endgroup$ – James Warthington Commented Oct 12, 2019 at 19:21
• 2 $\begingroup$ you're differentiating with respect to x, not n, my man. The variable in the limit is N not x. $\endgroup$ – astralwolf Commented Dec 23, 2019 at 19:45

| Show 1 more comment 35 $\begingroup$

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 $\mathrm{e}^x$.

Share Cite Follow edited Aug 20, 2016 at 19:11 answered Jul 30, 2014 at 15:39 Fly by NightFly by Night 32.7k44 gold badges5656 silver badges103103 bronze badges $\endgroup$ 5

• 16 $\begingroup$ what if there are infinite terms and then you cannot exchange the sum and limit if n is non integer $\endgroup$ – happymath Commented Mar 3, 2015 at 10:17
• 3 $\begingroup$ it's not easy to make this argument rigorous. This answer would be better if it pointed out that fact $\endgroup$ – wlad Commented May 30, 2019 at 18:28
• 4 $\begingroup$ a rigorous version of this argument uses the monotone convergence theorem $\endgroup$ – wlad Commented May 30, 2019 at 18:41
• $\begingroup$ rigorous version: math.stackexchange.com/a/1898375/445105 $\endgroup$ – Felix Benning Commented Sep 17, 2021 at 10:03
• $\begingroup$ @happymath $n$ is Never Non Ninteger :-) $\endgroup$ – Oskar Limka Commented Oct 12, 2021 at 17:08

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