How do I expand this exponential limit using Taylor expansion?
I have this equation:
$$\lim_{n\to \infty}\left(1+{x\over n} \right)^n=e^x$$
I need to Taylor expand both sides to prove that they are equal. Now I know how to do this using l'Hopital's rule, but I'm not supposed to prove this in that way. This is what I've tried so far:
$$\lim_{n\to \infty}\left(1+{x\over n} \right)^n$$ $$=\lim_{n\to \infty}e^{n\big[\log\big( 1+{x\over n}\big)\big]}$$ $$=\lim_{n\to \infty}\big(1+n\log\big(1+{x\over n}\big)+{n^2\over 2!}{\log^2\big(1+{x\over n}}\big)+O(x^3)\big)$$
This doesn't look anything like the expansion for $e^x$, so I tried to pass the limit through the exponential, expand $\log_{n\to \infty}$, and solve it that way:
$$e^{\lim_{n\to \infty}\big(1+{x\over n} \big)^n}$$ $$e^{\lim_{n\to \infty}\big[1+n\big({x\over n}-{x^2\over 2n^2}+{x^3\over 3n^3}-O(x^4)\big) \big]}$$
Distributing the $n$ into the parentheses, and taking the limit of my result was this:
$$e^{\lim_{n\to \infty}\big(1+{x\over n} \big)^n} = e^{1+x}=e^1e^x$$
And it looks very close to what I need it to be. I also have the problem that there aren't any factorials in the denominators of my expansion. I found this stack exchange question: Limit Question involving logarithmic taylor expansion. However, it didn't help me very much.
How should I expand this equation and what am I doing wrong?
$\endgroup$ 13 Answers
$\begingroup$By the binomial theorem, $$ \left( 1 + \frac{x}{n} \right)^n = \sum_{k=0}^n \frac{1}{n^k}\binom{n}{k} x^k = \sum_{k=0}^{\infty} \frac{1}{n^k}\binom{n}{k} x^k, $$ since the binomial coefficient is zero for $k>n$.
Meanwhile, the Taylor expansion of $e^x$ is supposed to be $$ \sum_{k=0}^{\infty} \frac{1}{k!}x^k. $$ Equating powers of $x$, we want to show that $$ \lim_{n \to \infty} \frac{1}{n^k} \binom{n}{k} = \frac{1}{k!}, $$ or expanding out the definition of the binomial coefficient, $$ \lim_{n \to \infty} \frac{n(n-1)\dotsm (n-k+1)}{n^k} = 1. $$ But this is straightforward: $$ \frac{n(n-1)\dotsm (n-k+1)}{n^k} = 1 \left( 1-\frac{1}{n} \right)\left( 1-\frac{2}{n} \right) \dotsm \left( 1-\frac{k+1}{n} \right), $$ and the latter is a polynomial in $1/n$, with constant term $1$, so it tends to $1$ as required.
$\endgroup$ $\begingroup$In fact with taylor expansion of $\ln(1+x)$ at $x=0$ you have $$ \ln(1+x)=x-\frac{x^2}{2}+o\left(x^2\right) $$ So with $\frac{x}{n}\underset{n \rightarrow +\infty}{\rightarrow}0$ you can use it and
$\endgroup$ $\begingroup$$$e^{n\ln\left(1+\frac{x}{n}\right)}=e^{n\left(\frac{x}{n}+o\left(\frac{1}{n}\right)\right)}=e^{x+o\left(1\right)}\underset{n \rightarrow +\infty}{\rightarrow}e^{x}$$
Taylor expansion on the RHS
$e^x = 1 + x + \frac 12 x^2 + \frac {1}{3!} x^3 + \cdots$
or
$e^x = \sum_\limits{n=0}^\infty \frac {x^n}{n!}$
binomial expansion on the left-hand side.
$(1+\frac {x}{n})^n =$$ 1 + x + \frac {n-1}{n}x^2 + \frac {(n-1)(n-2)}{n^2} x^3 + \cdots + \frac {(n-1)!}{n^{n-1}} x^n\\ 1 + x + (1-\frac {1}{n})x^2 + (1-\frac {1}{n})(1-\frac {2}n) x^3 + (1-\frac{1}{n})\cdots(1-\frac {n-1}n)x^n$
Every coefficient of $x^n$ on the left-hand side is less than or equal to the corresponding coefficient on the right-hand side.
So we can say for any $n$:
$(1+\frac {x}{n})^n \le e^x$
Choose $m<n$ Consider the first $m$ terms of the expansion of $(1+\frac {x}{n})^n$
Since each term is greater than 0
$ 1 + x + (1-\frac {1}{n})x^2 + (1-\frac {1}{n})(1-\frac {2}n) x^3 + (1-\frac{1}{n})\cdots(1-\frac {m-1}n)x^m \le (1+\frac xn)^n$
Now take the limits as $n\to \infty$ of both sides.
$\sum_\limits{n=1}^m \frac {x^n}{m}\le \lim_\limits{n\to \infty} (1+\frac xn)^n$
So:
$\sum_\limits{n=1}^m \frac {x^n}{n!} \le \lim_\limits{n\to \infty} (1+\frac xn)^n \le \sum_\limits{n=1}^\infty \frac {x^n}{n!}$
Allow $m$ to become large and by the squeeze theorem
$\lim_\limits{n\to \infty} (1+\frac xn)^n = e^x$
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