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The following first equality is intuitively evident: $$ \lim_{n\to \infty} \left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n = \lim_{n\to \infty} \left(1+\frac{x+y}{n}\right)^n =e^{x+y} $$ how can I give a formal argument?

Informally it is something like $$ \lim_{n\to \infty} \left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n =\lim_{n\to \infty} \left(1+\frac{\lim_{n\to \infty}\left(x+y+\frac{xy}{n}\right)}{n}\right)^n =\lim_{n\to \infty} \left(1+\frac{x+y}{n}\right)^n $$ Is there a "classical" property of sequences that backs this argument? Something like limit of composition, but here we speak of sequence only: $$ (b_n),\quad b_n \to b\quad\text{ for }n\to\infty \implies \lim_{n\to \infty} \left(1+\frac{b_n}{n}\right)^n =\lim_{n\to \infty} \left(1+\frac{b}{n}\right)^n $$

EDIT As Qiaochu Yuan is important to define what I have already defined. This computation is at the beginning of "Real Analysis", lets say in the chapter of sequences. Later, in the notes of complex analysis, I develop formal series with derivation and everything is the standard way (I think).

By this question I tried another approach, very early (1) define $$ e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n $$ (2) derive the product as depicted in this question (3) derive the derivation of the $e^x$.

Thus I have (a) no logarithm, (b) no derivation of $e^x$ yet.

Joaquin
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  • This is already handled in this answer but a simpler approach is the lemma of Thomas Andrews : if $n(a_n-1)\to 0$ then $a_n^n\to 1$. Choose $a_n=(1+(x/n)) (1+(y/n))/(1+((x+y)/n))$. – Paramanand Singh Jan 09 '21 at 04:05
  • @ParamanandSingh Thank you. The linked answers is the type of computation that originally I was aiming to. +1 – Joaquin Jan 10 '21 at 03:39

3 Answers3

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You can use the basic fact that for any sequence $c_n \stackrel{n\to\infty}{\longrightarrow}0$ $(c_n \neq 0)$ you have

$$\left(1+c_n\right)^{\frac 1{c_n}}\stackrel{n\to\infty}{\longrightarrow}e$$

So with $c_n =\frac{b_n}n$ and $b_n \stackrel{n\to\infty}{\longrightarrow}b$, you get

\begin{eqnarray*} \left(1+c_n\right)^n & = & \left(\left(1+c_n\right)^\frac 1{c_n}\right)^{c_n\cdot n}\\ & = & \left(\left(1+c_n\right)^\frac 1{c_n}\right)^{b_n} \\ & \stackrel{n\to\infty}{\longrightarrow} & e^b \end{eqnarray*}

Here the continuity of $f(x,y) = x^y$ in both variables $x$ and $y$ is used which can easily be established considering,for example, first the logarithm of this function.

Hence, you have

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

trancelocation
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  • This is nice. For the OP's purposes it's worth noting formally that we need joint continuity of the exponential $x^y$ in the base and the exponent here, which is true but may need to be established. – Qiaochu Yuan Jan 08 '21 at 08:01
  • @QiaochuYuan You are totally right. But i was a bit hesitant about mentioning this. – trancelocation Jan 08 '21 at 08:03
  • Very nice +1. My biggest problem is that I've no $x^y$ yet. What I have is $x^n$ and $x^{1/m}$. How is $x^y$ normally defined? Something like $x^y = \lim_{n\to\infty} x^{j_n/k_n}$ with $j_n/k_n\to y$, $j_n,k_n\in\mathbb{N}$? – Joaquin Jan 08 '21 at 08:17
  • @Joaquin: yes, that's one approach, I think the one taken in Rudin although it's been awhile. You can also define $x^y = e^{y \log x}$ if logarithms are available. – Qiaochu Yuan Jan 08 '21 at 08:20
  • @QiaochuYuan I see, the second is the one used by Apostol, which I follow the most. But this time I'll go with Rudin – Joaquin Jan 08 '21 at 08:26
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    Since you have $x^n$ and $x^{\frac 1m}$, you also have $x^q$ for rational $q$ and real non-negative $x$. So, a typical but quite technical way wrt. the proofs is defining for real $y=\lim_{n\to \infty}q_n$ for rational sequences:

    $$x^y :=\lim_{n\to\infty}x^{q_n}$$ This requires to show the independence of the sequence $q_n$. In a next step one would show that the 'limit rules' also hold for real sequences $x_n\stackrel{n\to\infty}{\longrightarrow}x$, $y_n\stackrel{n\to\infty}{\longrightarrow}y$: $$\lim_{n\to\infty}x_n^{y_n}=x^y$$ But this means exactly showing the continuity of $x^y$.

     – trancelocation Jan 08 '21 at 08:28
  • Perfect, this is the path I will follow – Joaquin Jan 08 '21 at 08:30
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The desired statement is true. It follows from taking logarithms and writing

$$\begin{eqnarray*} \log \left( 1 + \frac{b_n}{n} \right)^n &=& n \log \left( 1 + \frac{b_n}{n} \right) \\ &=& n \left( \frac{b_n}{n} + O \left( \frac{b_n^2}{n^2} \right) \right) \\ &=& b_n + O \left( \frac{b_n^2}{n} \right). \end{eqnarray*}$$

Taking the limit as $n \to \infty$ gives that the limit of the logarithms is $\lim_{n \to \infty} \left( b_n + O \left( \frac{b_n^2}{n} \right) \right) = b$ (using that $b_n$ is convergent and hence bounded, so the $O \left( \frac{b_n^2}{n} \right)$ term vanishes in the limit).

Depending on how you've defined exponentials and logarithms and proven their properties this argument may be circular, since it depends on the first two terms of the Taylor series of the logarithm, and one of several ways to compute the Taylor series of the logarithm is to use $\log xy = \log x + \log y$ which is more or less equivalent to $e^{x+y} = e^x e^y$.

There are alternatives for proving that $e^{x + y} = e^x e^y$ which avoid having to prove this; for example, taking for granted that $\frac{d}{dx} e^x = e^x$ and $e^0 = 1$ (which together uniquely determine $e^x$ by the existence and uniqueness theorems for ODEs), we have first

$$\frac{d}{dx} (e^x e^{-x}) = e^x e^{-x} - e^x e^{-x} = 0$$

so $e^x e^{-x}$ is a constant, and plugging in $x = 0$ gives $e^x e^{-x} = 1$. Next,

$$\frac{d}{dx} \left( e^{x+y} e^{-x} e^{-y} \right) = e^{x+y} e^{-x} e^{-y} - e^{x+y} e^{-x} e^{-y} = 0$$

so $e^{x+y} e^{-x} e^{-y}$ is a constant in $x$, and plugging in $x = 0$ gives $e^y e^{-y} = 1$ and hence $e^{x+y} = e^x e^y$.

Qiaochu Yuan
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  • Thank you. You pointed out the main question, what do I have, I enhanced my question. Both use of log and of derivative are at this point not available. +1 as soon as I have 15pts – Joaquin Jan 08 '21 at 08:00
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Another approach: Note that
$$\left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n = \left(1+\frac x n\right)^n\left(1+\frac y n\right)^n.$$

Addendum

We have that \begin{align*} \left(\frac{1+\frac{x+y+\frac{xy}{n}}{n}}{1+\frac{x+y}{n}}\right)^n &= \left( \frac{\left(1+\frac x n\right)\left(1+\frac y n\right)}{1+\frac{x+y}{n}}\right)^n \longrightarrow \frac{e^x e^y}{e^{x+y}} \quad(n\to\infty). \end{align*} It remains to show that this limit is unity. See this answer, for example.

user26872
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