My original goal was to prove that
$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$
using the squeeze theorem as we haven't seen differentiability yet and thus I cannot use arguments such as Taylor series nor Bernoulli's theorem, nor can I use induction. For that I wanted to find a lower and upper bound for $e^x$ in order to apply the squeeze theorem.
For the upper bound I used the fact that $x^n\leq x^2$ for $-1\leq x\leq 1$ and $n\geq 2$ thus one has that
\begin{align*}e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n&=\lim_{n\to\infty}\sum_{k=0}^n{n\choose k}\frac{x^k}{n^k}\\
&=\lim_{n\to\infty}1+x+\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}\\
&\leq \lim_{n\to\infty}1+x+\sum_{k=2}^n{n\choose k}\frac{x^2}{n^k}\\
&= \lim_{n\to\infty}1+x+\left(\sum_{k=2}^n{n\choose k}\frac{1}{n^k}\right)\cdot x^2\\
&= \lim_{n\to\infty}1+x+\left(\left(1+\frac{1}{n}\right)^n-2\right)\cdot x^2\\
&= 1+x+\left(e-2\right)\cdot x^2
\end{align*}
I could now potentially bound $x^n\geq -x^2$ in the same interval and obtain the bound
\begin{align*}e^x\geq 1+x-\left(e-2\right)\cdot x^2
\end{align*}
but I am not happy with it as I know that Bernoulli's inequality is stronger and gives
\begin{align*}e^x\geq 1+x.
\end{align*}
For $x\in (0,1)$ it's rather trivial to prove as
$$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}1+x+\underbrace{{n\choose 2}\frac{x^2}{n^2}+\cdots+\frac{x^n}{n^n}}_{\geq 0}\geq 1+x$$
but for $x\in(-1,0)$ the same argument does not apply straightforwardly due to the changing signs. So I modified it as follows:
For $-1\leq x\leq 0$ one has that $x^3\leq x^n$ ($x^3$ is in particular negative)
\begin{align*}
e^x&=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}1+x+{n\choose 2}\frac{x^2}{n^2}+\sum_{k=3}^n{n\choose k}\frac{x^k}{n^k}\\
& \geq \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\sum_{k=3}^n{n\choose k}\frac{x^3}{n^k}\\
& = \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\left((1+\frac{1}{n})^n-\frac{n-1}{2n}-2\right)x^3\\
& \geq \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\left(e-\frac{n-1}{2n}-2\right)x^3\\
&= \lim_{n\to\infty}1+x+x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)
\end{align*}
Now we not that the cubic function $x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)$ has a double zero at $x=0$ and the remaining zero is at
$$x=-\frac{\frac{n-1}{2n}}{e-\frac{n-1}{2n}-2}\overset{n\to \infty}{\longrightarrow} -\frac{1/2}{e-1/2-2}\cong -2.29$$
thus for $n$ sufficiently large the last zero is to the left of $-1$ and hence the function $x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)$ is positive on $(-1,0)$ thus
\begin{align*}
e^x& \geq \lim_{n\to\infty}1+x+\underbrace{x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)}_{\geq 0,\quad x\in(-1,0)}\\
&\geq 1+x
\end{align*}
Since I wrote up this proof I ask: could please give it a look and tell me if there are any mistakes or if there is a shorter solution which I overlooked?
Many thanks in advance!
Best Answer
As regards your original goal, there is a shorter way. We have that $$e^x-1-x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n-1-x=\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}.$$ Hence, for $x\in [-1,1]$, $$|e^x-1-x|=\left|\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}\right|\leq x^2\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{|x|^{k-2}}{n^k}\\\leq x^2\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{1}{n^k} \leq x^2\sum_{k=2}^{\infty}\frac{1}{k!}\leq e x^2$$ and the given limit follows as $x\to 0$ by the squeeze theorem.
Along the same argument we show that for $x\in [-1,1]$, $$\left|e^x-\sum_{k=0}^{n}\frac{x^{k}}{k!}\right|<e|x|^{n+1}$$ which implies that $e^x=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$.
Bernoulli inequality for $-1<x<0$: $$e^x-1-x=\sum_{k=2}^{\infty}\frac{x^{k}}{k!}=\sum_{k=1}^{\infty}\frac{x^{2k}}{(2k)!}\underbrace{\left(1+\frac{x}{2k+1}\right)}_{\geq 0}\geq 0.$$