Vtyjkyuk

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

$$e:=lim_ntoinftyleft(1+frac1nright)^n$$

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

(1) Deduce that

$$e=lim_ntoinftyleft(1+frac1f(n)right)^f(n);,;;textas long as;;f(n)xrightarrow[ntoinfty]infty$$

and then from here ($,xneq0,$ , but this is only a light technicality)

$$left(1+fracxnright)^n=left[;left(1+frac1fracnxright)^fracnx;right]^xxrightarrow[ntoinfty]e^x$$

2) For $,x>0,$ , substitute $,mx=n,$ . Note that $,ntoinftyimplies mtoinfty,$ , and

$$left(1+fracxnright)^n=left(left(1+frac1mright)^mright)^xxrightarrow[ntoinftyiff mtoinfty]e^x$$

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

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

Fundamentallemma. For every sequence of complex numbers $w_n$ with a limit $w$ it is true that $$lim_n to infty Bigl(1 + fracw_nnBigr)^n = sum_k=0^infty fracw^kk!.$$ Proof. For every $varepsilon > 0$ and sufficiently large index $K$ we have the following estimations: $$sum_k=K^infty frac(k! < frac varepsilon 3 quadmboxandquad |w_n| le |w|+1.$$Therefore if $n ge K$ then $$left|Bigl(1 + fracw_nnBig)^n - exp w right| le sum_k=0^K-1 left|n choose kfracw_n^kn^k - fracw^kk!right| + sum_k=K^nnchoose k fracw_nn^k + sum_k=K^infty frac^kk!.$$ The third sum is smaller than $varepsilon / 3$ based on our assumptions. We can find an upper bound for the middle one using $$n choose k frac 1 n^k = frac1k! prod_i = 1^k-1 Bigl(1 - frac i n Bigr) le frac 1 k!.$$ Combining this with $|w_n| le |w| + 1$, $$sum_k=K^n n choose k fracw_nn^k < sum_k=K^n frac(k! < frac varepsilon 3$$ 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.

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

$$ exp(x) := sum_n=0^infty fracx^nn!$$

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 $[0,fracan]$, so by the mean value theorem we know there exists a $c in [0,fracan]$ with

$$f'(c) = frac log(1+ fracan ) - log(1) frac an - 0 $$
$$ Longrightarrow log[(1+fracan)^n] = fraca1+c$$
$$ Longrightarrow (1+fracan)^n = exp(fraca1+c)$$

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

• As $c in [0,fracan]$ and $fracan rightarrow 0$ as $n rightarrow infty$, by the squeeze theorem we get that $ c rightarrow 0$ as $n rightarrow infty$


• As $ c rightarrow 0$ as $n rightarrow infty$, $fraca1+c rightarrow a$ 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 $fraca1+c rightarrow a$ as $n rightarrow infty$

$$ exp(fraca1+c) rightarrow exp(a) $$
as $n rightarrow infty$. Thus we can conclude that

$$ lim_n to infty (1+fracan)^n = e^a$$

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

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
$$
beginalign
f'(x)&=lim_hto 0fracf(x+h)-f(x)h\
&=lim_hto 0fracln(x+h)-ln(x)h\
&=lim_hto 0frac1hlnfracx+hx\
&=lim_h to 0lnleft(fracx+hxright)^frac1h\
&=lim_hto 0lnleft(1+frachxright)^frac1h
endalign
$$
Then, using the fact that $ln(x)$ is a continuous function for all $x$ in its domain, we can exchange the $lim$ and $ln$:
$$
f'(x)=lnlim_hto 0left(1+frachxright)^frac1h
$$
Now, let $m=1/h$. Then $mtoinfty$ as $hto 0^+$, and
$$
f'(x)=lnlim_mtoinftyleft(1+frac1mxright)^m
$$
Now, assuming $x>0$, define $n=mx^2$, and so $ntoinfty$ as $mtoinfty$. Then we can write
$$
f'(x)=lnlim_ntoinftyleft[left(1+fracxnright)^nright]^1/x^2
$$
and from before, we still have $f'(x)=1/x$, so
$$
lnlim_ntoinftyleft[left(1+fracxnright)^nright]^1/x^2=frac1x
$$
Exponentiating both sides, we find
$$
lim_ntoinftyleft[left(1+fracxnright)^nright]^1/x^2=e^1/x
$$
Finally, raising both sides to the $x^2$, we find
$$
lim_ntoinftyleft(1+fracxnright)^n=e^x
$$

Aaah... The sweet sound of silent revenge downvotes... Always a pleasure!

Consider the functions $u$ and $v$ defined for every $|t|ltfrac12$ by
$$
u(t)=t-log(1+t),qquad v(t)=t-t^2-log(1+t).
$$
The derivative of $u$ is $u'(t)=fract1+t$, which has the sign of $t$, hence $u(t)geqslant0$. The derivative of $v$ is $v'(t)=1-2t-frac11+t$, which has the sign of $(1+t)(1-2t)-1=-t(1+2t)$ which has the sign of $-t$ on the domain $|t|ltfrac12$ hence $v(t)leqslant0$.
Thus:

For every $|t|ltfrac12$,
$$
t-t^2leqslantlog (1+t)leqslant t.
$$

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

For every $|x|<frac12n$,
$$
expleft(x-fracx^2nright)leqslantleft(1+fracxnright)^nleqslantmathrm e^x.
$$

Finally, $x^2/nto 0$ when $ntoinfty$ and the exponential is continuous at $0$, hence we are done.

Facts/Definitions used:

• The logarithm has derivative $tmapsto1/t$.


• The exponential is the inverse of the logarithm.

$ (1+x/n)^n = sum_k=0^n binomnkfracx^kn^k $

Now just prove that $binomnkfracx^kn^k$ approaches $fracx^kk!$ as n approaches infinity, and you will have proven that your limit matches the Taylor series for $exp(x)$

For any fixed value of $x$, define

$$f(u)= ln(1+ux)over u$$

By L'Hopital's Rule,

$$lim_urightarrow0^+f(u)=lim_urightarrow0^+x/(1+ux)over1=x$$

Now exponentiate $f$:

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

By continuity of the exponential function, we have

$$lim_urightarrow0^+(1+ux)^1/u=lim_urightarrow0^+e^f(u)=e^lim_urightarrow0^+f(u)=e^x$$

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:

$$lim_nrightarrowinftyleft(1+xover nright)^n = e^x$$

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:

$log(lim_n(1 + fracxn)^n) = lim_n n log(1 + fracxn) = lim_n n cdot[fracxn - fracx^22n^2 + dots] = x$

EDIT: The logarithm is defined as usual: $log x = int_1^x fracdtt$. 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 $log(1+x)$.

There is at most one function $g$ on $mathbbR$ such that
$$g'(x)=g(x)text for all xtext in mathbbRquadtextandquad g(0)=1,.$$
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 criteria is what you should probably have in mind when you think of "the exponential".