I want to prove that for $m→∞ ⇒((1+x/m)^m→exp(x))$
My idea is to prove that there is an $m≥n$ so that $$(1+x/m)^m≥\sum_{j=0}^{n}x^j/j!≥(1+x/n)^n$$
now I would use the binomial theorem to rewrite $(1+x/m)^m, (1+x/n)^n$ as a sum, but I don't know how to prove both inequalities. With the left inequality I simply don't know which $m$ to choose, and in the right inequality I just get horribly stuck with the summations. Any help please?
To show
$$\lim_{n \to \infty}\left(1 + \frac{x}{n} \right)^n = e^x,$$
note that for $0 <y < 1$ we have
$$1 + y < e^y = \sum_{j=0}^\infty \frac{y^j}{j!} < \sum_{j=0}^\infty y^j = \frac{1}{1-y},$$
and for $y = x/n $ with $n > x$,
$$1 + \frac{x}{n} < e^{x/n} < \left(1 - \frac{x}{n} \right)^{-1}, \\ \implies \left(1 + \frac{x}{n} \right)^n < e^{x} \\ \implies \left(1 - \frac{x}{n} \right)^{n} < e^{-x} $$
Hence, for $x \in \mathbb{R}$,
$$\begin{align}0 &\leqslant e^x - \left(1 + \frac{x}{n} \right)^n \\ &= e^x \left[1 - e^{-x} \left(1 + \frac{x}{n} \right)^n\right] \\ &< e^x \left[1 - \left(1 - \frac{x^2}{n^2} \right)^n\right] .\end{align}.$$
Using Bernoulli's inequality we have
$$\left(1 - \frac{x^2}{n^2} \right)^n > 1 - \frac{x^2}{n}.$$
Hence,
$$0 \leqslant e^x - \left(1 + \frac{x}{n} \right)^n< \frac{e^xx^2}{n},$$
and the desired limit follows from the squeeze theorem.
There are other proofs on MSE, for example this one, but I find its use of the floor function unnecessarily fiddly. I think the following is a bit more straightforward, but a certain amount of grunge seems unavoidable.
Let's prove that difference between the two expressions becomes small as $n \to \infty$. In other words, given $\epsilon > 0$, we want to show that there is some $N$ such that whenever $n \geq N$, we have $$e_n(x) = \left|\left(1 + \frac{x}{n}\right)^n - \sum_{k=0}^{n}\frac{x^k}{k!}\right| < \epsilon$$ We manipulate the right hand side as follows: $$\begin{aligned} e_n(x) &= \left|\sum_{k=0}^{n}{n \choose k}\left(\frac{x}{n}\right)^k - \sum_{k=0}^{n}\frac{x^k}{k!}\right|\\ &= \left|\sum_{k=0}^{n} \frac{x^k}{k!} \left( \frac{n!}{(n-k)!n^k} - 1\right)\right| \\ &= \left|\sum_{k=0}^{n} \frac{x^k}{k!} \left( \frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdots\frac{n-k+1}{n} - 1\right)\right| \\ &= \left|\sum_{k=0}^{n} \frac{x^k}{k!} \left( 1 \cdot \left(1 - \frac{1}{n}\right) \cdot \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{k-1}{n}\right) - 1\right)\right| \\ &\leq \sum_{k=0}^{n}\frac{|x|^k}{k!} \left|a_{k,n} - 1\right| \end{aligned}$$ where $$a_{k,n} = \left(1 - \frac{1}{n}\right) \cdot \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{k-1}{n}\right)$$ Note that for $0 \leq k \leq n$, all of the factors are between $0$ and $1$, so $$0 < a_{k,n} \leq 1$$ Let $\epsilon > 0$.
For fixed $k$, it is clear that $a_{k,n} \to 1$ as $n \to \infty$, so $|a_{k,n} - 1| \to 0$ as $n \to \infty$. Therefore, there is some $N_k$ such that $|a_{k,n} - 1| < \epsilon$ for all $n \geq N_k$.
Now, fix a positive integer $m$. If we set $N = \max\{N_1,N_2,\ldots,N_m\}$, then $|a_{k,n} - 1| < \epsilon$ for all $k \leq m$ and $n \geq N$, in which case $$\begin{aligned} e_n(x) &\leq \sum_{k=0}^{m}\frac{|x|^k}{k!}|a_{k,n}-1| + \sum_{k=m+1}^{n}\frac{|x|^k}{k!}|a_{k,n}-1| \\ &\leq \epsilon \sum_{k=0}^{m}\frac{|x|^k}{k!} + \sum_{k=m+1}^{n}\frac{|x|^k}{k!} \\ &\leq \epsilon \sum_{k=0}^{\infty}\frac{|x|^k}{k!} + \sum_{k=m+1}^{\infty}\frac{|x|^k}{k!} \\ &= \epsilon e^{|x|} + \sum_{k=m+1}^{\infty}\frac{|x|^k}{k!} \\ \end{aligned}$$ Therefore, $$\limsup_{n \to \infty}e_n(x) \leq \epsilon e^{|x|} + \sum_{k=m+1}^{\infty}\frac{|x|^k}{k!}$$ Letting $m \to \infty$ we get $$\limsup_{n \to \infty}e_n(x) \leq \epsilon e^{|x|}$$ As this is true for any $\epsilon > 0$, we conclude that $\limsup_{n \to \infty}e_n(x) = 0$, and therefore the limit exists and $$\lim_{n \to \infty}e_n(x) = 0$$ as desired.
