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

Solution 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.

Solution 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)

Solution 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)

Solution 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:

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