About $\lim \left(1+\frac {x}{n}\right)^n$

Question

I was wondering if it is possible to get a link to a rigorous proof that $$\displaystyle \lim_{n\to\infty} \left(1+\frac {x}{n}\right)^n=\exp x$$

Answer 1

I would like to cite here an awesome German mathematician, Konrad Königsberger. He writes in his book Analysis I as follows:

Fundamental lemma. For every sequence of complex numbers $w_n$ with a limit $w$ it is true that $$\lim_{n \to \infty} \Bigl(1 + \frac{w_n}{n}\Bigr)^n = \sum_{k=0}^\infty \frac{w^k}{k!}.$$ Proof. For every $\varepsilon > 0$ and sufficiently large index $K$ we have the following estimations: $$\sum_{k=K}^\infty \frac{(|w|+1)^k}{k!} < \frac \varepsilon 3 \quad\mbox{and}\quad |w_n| \le |w|+1.$$Therefore if $n \ge K$ then $$\left|\Bigl(1 + \frac{w_n}{n}\Big)^n - \exp w \right| \le \sum_{k=0}^{K-1} \left|{n \choose k}\frac{w_n^k}{n^k} - \frac{w^k}{k!}\right| + \sum_{k=K}^n{n\choose k} \frac{|w_n|^k}{n^k} + \sum_{k=K}^\infty \frac{|w|^k}{k!}.$$ The third sum is smaller than $\varepsilon / 3$ based on our estimations. We can find an upper bound for the middle one using $${n \choose k} \frac 1 {n^k} = \frac{1}{k!} \prod_{i = 1}^{k-1} \Bigl(1 - \frac i n \Bigr) \le \frac 1 {k!}.$$ Combining this with $|w_n| \le |w| + 1$, $$\sum_{k=K}^n {n \choose k} \frac{|w_n|^k}{n^k} < \sum_{k=K}^n \frac{(|w|+1)^k}{k!} < \frac \varepsilon 3$$ Finally, the first sum converges to $0$ due to $w_n \to w$ and ${n \choose k} n^{-k} \to \frac 1 {k!}$. We can choose $N > K$ such that it's smaller than $\varepsilon / 3$ as soon as $n > N$.

Really brilliant.

Answer 2

From the very definition (one of many, I know):

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

we can try the following, depending on what you have read so far in this subject:

(1) Deduce that

$$e=\lim_{n\to\infty}\left(1+\frac{1}{f(n)}\right)^{f(n)}\;,\;\;\text{as long as}\;\;f(n)\xrightarrow[n\to\infty]{}\infty$$

and then from here ($\,x\neq0\,$ , but this is only a light technicality)

$$\left(1+\frac{x}{n}\right)^n=\left[\;\left(1+\frac{1}{\frac{n}{x}}\right)^\frac{n}{x}\;\right]^x\xrightarrow[n\to\infty]{}e^x$$

2) For $\,x>0\,$ , substitute $\,mx=n\,$ . Note that $\,n\to\infty\implies m\to\infty\,$ , and

$$\left(1+\frac{x}{n}\right)^n=\left(\left(1+\frac{1}{m}\right)^m\right)^x\xrightarrow[n\to\infty\iff m\to\infty]{}e^x$$

I'll leave it to you to work out the case $\,x<0\,$ (hint: arithmetic of limits and "going" to denominators)

Answer 3

Firstly, let us give a definition to the exponential function, so we know the function has various properties:

$$ \exp(x) := \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

so that we can prove that (as exp is a power series) :

• The exponential function has radius of convergence $\infty$, and is thus defined on all of $\mathbb R$
• As a power series is infinitely differentiable inside its circle of convergence, the exponential function is infinitely differentiable on all of $\mathbb R$
• We can then prove that the function is strictly increasing, and thus by the inverse function theorem (http://en.wikipedia.org/wiki/Inverse_function_theorem) we can define what we know as the "log" function

Knowing all of this, here is hopefully a sufficiently rigorous proof (at least for positive a):

As $\log(x)$ is continuous and differentiable on $(0,\infty)$, we have that $\log(1+x)$ is continuous and differentiable on $[0,\frac{a}{n}]$, so by the mean value theorem we know there exists a $c \in [0,\frac{a}{n}]$ with

$$f'(c) = \frac {\log(1+ \frac{a}{n} ) - \log(1)} {\frac {a}{n} - 0 } $$ $$ \Longrightarrow \log[{(1+\frac{a}{n})^n}] = \frac{a}{1+c}$$ $$ \Longrightarrow (1+\frac{a}{n})^n = \exp({\frac{a}{1+c}})$$

for some $c \in [0,\frac{a}{n}]$ . As we then want to take the limit as $n \rightarrow \infty$, we get that:

• As $c \in [0,\frac{a}{n}]$ and $\frac{a}{n} \rightarrow 0$ as $n \rightarrow \infty$, by the squeeze theorem we get that $ c \rightarrow 0$ as $n \rightarrow \infty$
• As $ c \rightarrow 0$ as $n \rightarrow \infty$, $\frac{a}{1+c} \rightarrow a$ as $n \rightarrow \infty$
• As the exponential function is continuous on $\mathbb R$, the limit can pass inside the function, so we get that as $\frac{a}{1+c} \rightarrow a$ as $n \rightarrow \infty$

$$ \exp(\frac{a}{1+c}) \rightarrow \exp(a) $$ as $n \rightarrow \infty$. Thus we can conclude that

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

(Of course, this is ignoring that one needs to prove that $\exp(a)=e^a$, but this is hardly vital for this question)

Answer 4

For any fixed value of $x$, define

$$f(u)= {\ln(1+ux)\over u}$$

By L'Hopital's Rule,

$$\lim_{u\rightarrow0^+}f(u)=\lim_{u\rightarrow0^+}{x/(1+ux)\over1}=x$$

Now exponentiate $f$:

$$e^{f(u)}=(1+ux)^{1/u}$$

By continuity of the exponential function, we have

$$\lim_{u\rightarrow0^+}(1+ux)^{1/u}=\lim_{u\rightarrow0^+}e^{f(u)}=e^{\lim_{u\rightarrow0^+}f(u)}=e^x$$

All these limits have been shown to exist for the (positive) real variable $u$ tending to $0$, hence they must exist, and be the same, for the sequence of reciprocals of integers, $u=1/n$, as $n$ tends to infinity, and the result follows:

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

Answer 5

Another answer, assuming $x>0$:

Let $f(x)=\ln(x)$. Then we know that $f'(x)=1/x$. Also, by the definition of derivative, we can write $$ \begin{align} f'(x)&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\to 0}\frac{\ln(x+h)-\ln(x)}{h}\\ &=\lim_{h\to 0}\frac{1}{h}\ln\frac{x+h}{x}\\ &=\lim_{h \to 0}\ln\left(\frac{x+h}{x}\right)^\frac{1}{h}\\ &=\lim_{h\to 0}\ln\left(1+\frac{h}{x}\right)^\frac{1}{h} \end{align} $$ Then, using the fact that $\ln(x)$ is a continuous function for all $x$ in its domain, we can exchange the $\lim$ and $\ln$: $$ f'(x)=\ln\lim_{h\to 0}\left(1+\frac{h}{x}\right)^\frac{1}{h} $$ Now, let $m=1/h$. Then $m\to\infty$ as $h\to 0^+$, and $$ f'(x)=\ln\lim_{m\to\infty}\left(1+\frac{1}{mx}\right)^m $$ Now, assuming $x>0$, define $n=mx^2$, and so $n\to\infty$ as $m\to\infty$. Then we can write $$ f'(x)=\ln\lim_{n\to\infty}\left[\left(1+\frac{x}{n}\right)^n\right]^{1/x^2} $$ and from before, we still have $f'(x)=1/x$, so $$ \ln\lim_{n\to\infty}\left[\left(1+\frac{x}{n}\right)^n\right]^{1/x^2}=\frac{1}{x} $$ Exponentiating both sides, we find $$ \lim_{n\to\infty}\left[\left(1+\frac{x}{n}\right)^n\right]^{1/x^2}=e^{1/x} $$ Finally, raising both sides to the $x^2$, we find $$ \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x $$ EDIT: This idea actually works for all reals—if we use $f(x)=\ln|x|$ instead, then we get eventually get: $$ e^x=\lim_{n\to\infty}\left|1+\frac{x}{n}\right|^{n}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n $$ Where the last equality come from the fact that $n$ always eventually dominates $x$, so that the absolute value function becomes redundant.

This leaves the case where $x=0$, but that is a trivial matter.

Answer 6

Consider the functions $u$ and $v$ defined for every $|t|\lt\frac12$ by $$ u(t)=t-\log(1+t),\qquad v(t)=t-t^2-\log(1+t). $$ The derivative of $u$ is $u'(t)=\frac{t}{1+t}$, which has the sign of $t$, hence $u(t)\geqslant0$. The derivative of $v$ is $v'(t)=1-2t-\frac{1}{1+t}$, which has the sign of $(1+t)(1-2t)-1=-t(1+2t)$ which has the sign of $-t$ on the domain $|t|\lt\frac12$ hence $v(t)\leqslant0$. Thus:

For every $|t|\lt\frac12$, $$ t-t^2\leqslant\log (1+t)\leqslant t. $$

The function $z\mapsto\exp(nz)$ is nondecreasing on the same domain hence $$ \exp\left(nt-nt^2\right)\leqslant(1+t)^n\leqslant\exp\left(nt\right). $$ In particular, using this for $t=x/n$, one gets:

For every $|x|<\frac12n$, $$ \exp\left(x-\frac{x^2}{n}\right)\leqslant\left(1+\frac{x}n\right)^n\leqslant\mathrm e^x. $$

Finally, $x^2/n\to 0$ when $n\to\infty$ and the exponential is continuous at $0$, hence we are done.

Facts/Definitions used:

• The logarithm has derivative $t\mapsto1/t$.
• The exponential is the inverse of the logarithm.

Answer 7

Let $t\in[1,1+\frac{1}{n}]$, and we have: $$\frac{1}{1+\frac{1}{n}}\leq \frac{1}{t}\leq 1$$ $$\int_1^{1+\frac{x}{n}}\frac{1}{1+\frac{1}{n}}dt\leq \int_1^{1+\frac{x}{n}}\frac{1}{t}\leq \int_1^{1+\frac{x}{n}} 1dt$$ $$\frac{x}{n+1}\leq \ln\Big(1+\frac{x}{n}\Big)\leq\frac{x}{n}$$ Since $e^x$ is increasing on $\mathbb{R}$, we have: $$e^{\frac{x}{n+1}}\leq 1+\frac{x}{n}\leq e^{\frac{x}{n}}$$ Since $x^n$ is increasing on $(0,\infty)$, on the RHS we have: $$\Big(1+\frac{x}{n}\Big)^n\leq e^x$$ On the LHS we have: $$e^x\leq \Big(1+\frac{x}{n}\Big)^{n+1} \ \ \Longrightarrow \ \ \frac{e^x}{1+\frac{x}{n}}\leq \Big(1+\frac{x}{n}\Big)^n$$ Then we have: $$\frac{e^x}{1+\frac{x}{n}}\leq \Big(1+\frac{1}{n}\Big)^n\leq e^x$$ As $n\to\infty$ we know $\Big\{\displaystyle\frac{e^x}{1+\frac{x}{n}}\Big\}\to e^x$. So, by Squeeze Theorem, we have proved the limit.

Answer 8

$ (1+x/n)^n = \sum_{k=0}^n \binom{n}{k}\frac{x^k}{n^k} $

Now just prove that $\binom{n}{k}\frac{x^k}{n^k}$ approaches $\frac{x^k}{k!}$ as n approaches infinity, and you will have proven that your limit matches the Taylor series for $\exp(x)$

Answer 9

Using a single interval upper and lower Riemann sum for $\ln x:=\int_1^x\dfrac 1t\operatorname dt$, we get $\dfrac x{n+x}\le\ln(1+\frac xn)\le\dfrac xn$.

So, $e^{\frac x{n+x}}\le 1+\dfrac xn\le e^{\frac xn}$.

Now $e^{\frac {nx}{n+x}}\le (1+\dfrac xn)^n\le e^ x$.

Let $n\to\infty $.

I think I saw this in Best and Penner's Calculus.

Answer 10

This one of the ways in which it is defined. The equivalence of the definitions can be proved easily, I guess. If for example you take the exponential function to be the inverse of the logarithm:

$\log(\lim_n(1 + \frac{x}{n})^n) = \lim_n n \log(1 + \frac{x}{n}) = \lim_n n \cdot[\frac{x}{n} - \frac{x^2}{2n^2} + \dots] = x$

EDIT: The logarithm is defined as usual: $\log x = \int_1^x \frac{dt}{t}$. The first identity follows from the continuity of the logarithm, the second it's just an application of one of the property of the logarithm ($\log a^b = b \log a $), while to obtain the third it sufficies to have the Taylor expansion of $\log(1+x)$.

Answer 11

There is at most one function $g$ on $\mathbb{R}$ such that $$g'(x)=g(x)\text{ for all } x\text{ in }\mathbb{R}\quad\text{and}\quad g(0)=1\,.$$ If you let $f_n(x)=(1+x/n)^n$ and you can demonstrate that it compactly converges to some function $f$, you can demonstrate that $f'(x)=f(x)$ and $f(0)=1$. Likewise, if you take $f_n(x)=\sum_{k=0}^n x^k/k!$ and demonstrate this sequence converges compactly, you can show that this limit satisfies the same conditions. Thus it doesn't matter what your definition is. The uniqueness criterion is what you should probably have in mind when you think of "the exponential".

Answer 12

Start with the binomial theorem: $$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}\sum_{m=0}^n\frac{n!}{m!(n-m)!}\frac{x^m}{n^m}=\lim_{n\to\infty}\sum_{m=0}^n\frac{n!}{n^m(n-m)!}\frac{x^m}{m!}.$$ Define $$f(\mu,\nu)=\sum_{m=0}^{\nu}\frac{\mu!}{\mu^m(\mu-m)!}\frac{x^m}{m!}.$$ Thus you need to prove $$\lim_{n\to\infty}f(n,n)=\lim_{n\to\infty}\sum_{m=0}^n\frac{x^m}{m!},$$ which amounts to proving that $$\lim_{n\to\infty}f(n,n)=\lim_{\nu\to\infty}\left[\lim_{\mu\to\infty}f(\mu,\nu)\right]$$ and $$\forall{m\in\mathbb{N}},\lim_{\mu\to\infty}\frac{\mu!}{\mu^m(\mu-m)!}=1.$$ The latter can be proven rather easily using the principle of induction. The former follows immediately from the Tannery's theorem. With this done, we have that $$\lim_{n\to\infty}\sum_{m=0}^n\frac{n!}{n^m(n-m)!}\frac{x^m}{m!}=\lim_{\nu\to\infty}\left(\sum_{m=0}^{\nu}\left[\lim_{\mu\to\infty}\frac{\mu!}{\mu^m(\mu-m)!}\right]\frac{x^m}{m!}\right)=\lim_{\nu\to\infty}\sum_{m=0}^{\nu}\frac{x^m}{m!}:=\exp(x).$$ Even if the definition of $\exp$ you are using is not via its Maclaurin series, any other definition can be easily proven from Maclaurin series rather easily.

Answer 13

Not a rigorous but still a proof.

$\displaystyle \lim_{n \to \infty} \left(1+ \frac{x}{n} \right)^n = [1^ \infty]$ - uncertainty

Let $\displaystyle \lim_{x \to x_o}u(x)=1$ and $\displaystyle \lim_{x\to x_o}V(x)=\infty$, then

$\displaystyle \lim_{x \to x_o}u^V=\lim_{x \to x_o}\Big[ \Big(1+(u-1) \Big)^{\frac{1}{u-1}}\space \Big]^{(u-1)V}=e^{\displaystyle\lim_{x\to x_o}(u-1)V}$

Now let's apply what we got above to our problem. Here $ \{ u(n) = 1+\frac{x}{n}\} \to 1 $ and $\{V(n)=n\} \to \infty$ as $n\to \infty$.

$\displaystyle \lim_{n \to \infty} \left(1+ \frac{x}{n} \right)^n = [1^ \infty] = \exp \left[ \lim_{n\to \infty}\left(1+\frac{x}{n}-1 \right)n \right] = \exp \left[ \lim_{n\to \infty} \left( \frac{x}{n} \cdot n \right) \right] = \exp \left[ \lim_{n\to \infty}x \right] = \exp (x) = e^x$