How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion? Is it possible to determine the limit
$$\lim_{x\to0}\frac{e^x-1-x}{x^2}$$
without using l'Hopital's rule nor any series expansion?
For example, suppose you are a student that has not studied derivative yet (and so not even Taylor formula and Taylor series).
 A: Let us call our limit $\ell$.
I was considering the following identity
$$
4\frac{e^{2x}-1-2x}{(2x)^2}-2\frac{e^x-1-x}{x^2}=\left(\frac{e^x-1}{x}\right)^2\quad\forall x\ne0
$$
If $\mathbf{\ell}$ exists and is not infinite, taking the limit of the above identity we have
$$
4\ell-2\ell=1\implies\ell=\frac{1}{2}
$$
but I am not able to prove the bold part above (if at all possible, in a simple way).
A: Define $f(x)=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$.  One possibility is to take $f(x)$ as the definition of $e^x$.  Since the OP has suggested a different definition, I will show they agree.
If $x=\frac{p}{q}$ is rational, then
\begin{eqnarray*}
f(x)&=&\lim_{n\to\infty}\left(1+\frac{p}{qn}\right)^n\\
&=&\lim_{n\to\infty}\left(1+\frac{p}{q(pn)}\right)^{pn}\\
&=&\lim_{n\to\infty}\left(\left(1+\frac{1}{qn}\right)^n\right)^p\\
&=&\lim_{n\to\infty}\left(\left(1+\frac{1}{(qn)}\right)^{(qn)}\right)^{p/q}\\
&=&\lim_{n\to\infty}\left(\left(1+\frac{1}{n}\right)^{n}\right)^{p/q}\\
&=&e^{p/q}
\end{eqnarray*}
Now, $f(x)$ is clearly non-decreasing, so
$$
\sup_{p/q\leq x}e^{p/q}\leq f(x)\leq \inf_{p/q\geq x}e^{p/q}
$$
It follows that $f(x)=e^x$.
Now, we have
\begin{eqnarray*}
\lim_{x\to0}\frac{e^x-1-x}{x^2}&=&\lim_{x\to0}\lim_{n\to\infty}\frac{\left(1+\frac{x}{n}\right)^n-1-x}{x^2}\\
&=&\lim_{x\to0}\lim_{n\to\infty}\frac{n-1}{2n}+\sum_{k=3}^n\frac{{n\choose k}}{n^k}x^{k-2}\\
&=&\frac{1}{2}+\lim_{x\to0}x\lim_{n\to\infty}\sum_{k=3}^n\frac{{n\choose k}}{n^k}x^{k-3}\\
\end{eqnarray*}
We want to show that the limit in the last line is 0.  We have $\frac{{n\choose k}}{n^k}\leq\frac{1}{k!}\leq 2^{-(k-3)}$, so we have
\begin{eqnarray*}
\left|\lim_{x\to0}x\lim_{n\to\infty}\sum_{k=3}^n\frac{{n\choose k}}{n^k}x^{k-3}\right|&\leq&\lim_{x\to0}|x|\lim_{n\to\infty}\sum_{k=3}^n \left(\frac{|x|}{2}\right)^{k-3}\\
&=&\lim_{x\to0}|x| \frac{1}{1-\frac{|x|}{2}}\\
&=&0
\end{eqnarray*}
A: 
I thought it might be useful to present a way forward that relies on an integral representation of the numerator along with the mean-value theorem for integrals.  To that end, we now proceed.


Note that we can write the numerator as
$$\begin{align}
e^x-x-1&=\int_0^x \int_0^t e^s \,ds\,dt\\\\
&=\int_0^x \int_s^x e^s\,dt\,ds\\\\
&=\int_0^x (x-s)e^s\,ds
\end{align}$$
Next, we apply the Mean-Value-Theorem for integrals to reveal
$$\begin{align}
e^x-x-1&=e^{s^*}\int_0^x(x-s)\,ds\\\\
&=\frac12 x^2e^{s^*}
\end{align}$$
for some value of $s^*\in (0,x)$.
Finally, exploiting the continuity of the exponential function yields the coveted limit
$$\begin{align}
\lim_{x\to 0}\frac{e^x-x-1}{x^2}&=\lim_{x\to 0}\frac{\frac12 x^2e^{s^*}}{x^2}\\\\
&=\frac12
\end{align}$$
as expected!
A: Given the familiar limit ${e^t-1\over t}\to1$ as $t\to0$, we can argue as follows:
$${e^x-1-x\over x^2}={1\over x^2}\int_0^x(e^u-1)\,du={1\over x}\int_0^1(e^{xv}-1)\,dv=\int_0^1v\left(e^{xv}-1\over xv\right)\,dv$$
Now intuitively we have ${e^{xv}-1\over xv}={e^t-1\over t}\to1$ as $xv=t\to0$, hence
$$\int_0^1v\left(e^{xv}-1\over xv\right)\,dv\to\int_0^1v\cdot1\,dv={1\over2}v^2\Big|_0^1={1\over2}$$
but to be rigorous we need to justify the intuition of bringing the limit inside the integral. If you have a fancy enough theorem, you can cite it and be done, but let's do it from first principles: We need to show that for any $\epsilon\gt0$, there is a $\delta\gt0$ such that $0\lt|x|\lt\delta$ implies
$$\left|\int_0^1 v\left(e^{xv}-1\over xv\right)\,dv-{1\over2} \right|=\left|\int_0^1 v\left({e^{xv}-1\over xv}-1\right)\,dv\ \right|\le\epsilon$$
What we do know (from the familiar limit) is that for any $\epsilon\gt0$ there is a $\delta\gt0$ such that $0\lt|t|\lt\delta$ implies $\left|{e^t-1\over t}-1\right|\lt\epsilon$. Now if $0\lt|x|\lt\delta$ and $0\lt v\lt1$, then $0\lt|xv|\lt\delta$ as well, so for any $\epsilon\gt0$ we conclude there is a $\delta\gt0$ such that $0\lt|x|\lt\delta$ implies
$$\left|\int_0^1 v\left({e^{xv}-1\over xv}-1\right)\,dv\ \right|\le\int_0^1 v\left|{e^{xv}-1\over xv}-1\right|\,dv\
\le\int_0^1v\epsilon\,dv={\epsilon\over2}\lt\epsilon$$
and we're done.
Remark: The "familiar" limit ${e^t-1\over t}\to1$ as $t\to0$ is the definition of the derivative of the exponential function at $0$. The whole proof here is built around knowing that $(e^x)'=e^x$, in the form $\int_0^xe^u\,du=e^x-1$.
A: Accidentally I came across this post and I thought of
how to prove the statement
$$ \lim_{x\rightarrow 0} \frac{e^x-1-x}{x^2} = \frac12 .$$
assuming only
that the function $e^x$
satisfies the two properties:
$$ e^{x+y}=e^x e^y \ \mbox{and} \
 \lim_{x\rightarrow 0} \frac{e^x-1}{x} = 1$$
It turns out that it is possible using
$\sum_{k=0}^{n-1} = \frac{n(n-1)}{2}$ and elementary
algebraic properties of limits, but being very careful
with the uniform bounds for these limits.
The proof, although elementary,
is not simple so is probably not of much practical use.
Also, all the difficulties are hidden in the existence of
the function $e^x$ verifying the functional equation.
Anyway, I post it for the curios reader.
To start, note that
the second property for $e^x$ is equivalent to the following:
Write $R(x) = e^x-1-x$. Then
for $\delta>0$ and $|x|\leq \delta$ we have the uniform bound:
$|R(x)|\leq \Delta(\delta)$ with a function $\Delta$ that verifies:
$$\lim_{\delta\rightarrow 0}
\frac{\Delta(\delta)}{\delta} \rightarrow 0.$$
By the above definitions we also have
$|e^x|\leq M(\delta) \equiv  1+\delta+\Delta(\delta)
<+\infty$.
Fix $x\neq 0$, $L=\Delta(|x|)/|x|$, $M=M(|x|)$
and let $n\geq 1$. Using the functional equation
for $e^x$ we may rewrite $e^x-1=e^{nx/n}-1$ as a
telescopic sum:
$$ e^x-1=
\sum_{k=0}^{n-1} 
e^{\frac{k}{n} x} 
\left(
e^{\frac{x}{n}} 
-1\right)=
\sum_{k=0}^{n-1} 
\left( 1+ \frac{k}{n}x + R(\frac{k}{n}x) \right)
\left( \frac{x}{n} + R(\frac{x}{n}) \right)
$$
Developing the RHS
and using $\sum_{k=0}^{n-1} k = \frac{n^2-n}{2}$
we get the expression $x + \frac{n-1}{2n} x^2$
plus an error term which is bounded by
$$
\sum_{k=0}^{n-1} 
\left[ \Delta( \frac{k}{n}|x|) 
\times  (1+L) \frac{|x|}{n}+
e^{\frac{k}{n} x}  
 \Delta(\frac{|x|}{n}) \right] \leq 
n \Delta(|x|) 
\times  (1+L) \frac{|x|}{n}
 + M n  \times
 \Delta(\frac{|x|}{n})  $$
Therefore,
$$\left| \frac{e^x-(1+x+x^2/2)}{x^2} \right| \leq \frac{x^2}{2n} +
  (1+L) \frac{\Delta(|x|)}{|x|}  
 + M  \frac{1}{|x|} \frac{\Delta(|x|/n)}{|x|/n}  $$
Now let $n\rightarrow \infty$ (keeping $x\neq 0$ fixed).
By the properties of the function $\Delta$, the first and the last
terms on the RHS goes to zero and
as the LHS is independent of $n$ we deduce:
$$\left|\frac{e^x-(1+x+x^2/2)}{x^2} \right| \leq 
 (1+L(|x|)) \frac{\Delta(|x|)}{|x|}  .
$$
The RHS goes to zero as $|x|$ goes to zero,
and this implies the stated
limit.
Remark: Incidentally one may use the same telescopic procedure,
i.e. without binomial expansion,
to show that for $x$ fixed,
$e^x - (1+\frac{x}{n})^n \rightarrow 0$ as $n\rightarrow \infty$.
A: Consider fundamental limit: $e = \lim\limits_{n\to \infty}(1+\frac{1}{n})^n$ and $e^x = \lim\limits_{n\to\infty}(1+\frac{x}{n})^n$
Proof
$e^x = [\lim\limits_{k\to\infty}(1+1/k)^k]^x = \lim\limits_{k\to \infty}((1+1/k)^{kx})\Rightarrow kx = n
\Rightarrow e^x = \lim\limits_{n\to\infty}(1+\frac{x}{n})^n$.
Understand the first expression:
$P = \large\frac{e^x-1}{x}$
Note that $e^x - 1 - x = x.[\large\frac{(e^x-1)}{x} - 1]\,\,\therefore\,\,$ $\boxed{\lim\limits_{x\to 0}\frac{e^x-1-x}{x^2}=\lim\limits_{x\to 0}\frac{P-1}{x}}$
Lets go to understand the expression $\,\,P-1$.
$P - 1= \frac{e^x - 1}{x} - 1 = \lim\limits_{n\to\infty}\left(\large\frac{[(1+\frac{x}{n})^n - 1]}{x} - 1\right)=$
Using that tool:
$\boxed{b^n - 1 = (b-1).(b^{n-1}+b^{n-2}+...+1)}$
$=\lim\limits_{n\to\infty}\left((1+\frac{x}{n}-1).\large\frac{[(1+x/n)^{n-1} + (1+x/n)^{n-2} + ... + {1+x/n}]}{x}-1 \right) =\\
\\
= \lim\limits_{n\to\infty}\left(\frac{1}{n}.[(1+x/n)^{n-1} + (1+x/n)^{n-2} + ... + (1+x/n)]-1\right) = \\
\\
=\lim\limits_{n\to\infty}\frac{1}{n}.\left((1+x/n)^{n-1} + (1+x/n)^{n-2} + ... + (1+x/n)-n\right)$
Writing the last "$n$" as $\underbrace{1+1+1...+1}_{n\,\, times}$ and inputing these $1`s$ into it:
$P-1 = \lim\limits_{n\to\infty} (1/n).[((1+x/n)^{n-1} - 1)+ ((1+x/n)^{n-2} - 1) + ... + ((1+x/n) - 1)]$
Using again that tool in each expression:
$=\lim\limits_{n\to\infty}(\frac{1}{n}).(\frac{x}{n}) [((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... +1)+((1+x/n)^{n-3}+...+1)+...+1]$
Finally,
$L = \lim\limits_{x\to 0}\frac{P-1}{x} =\lim\limits_{x\to 0}\lim\limits_{n\to\infty}(\frac{1}{n}).(\frac{x}{n})[\large\frac{((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... + 1 ) + ( (1+x/n)^{n-3} + ... + 1 ) + ... + 1)}{x}]=$
$=\lim\limits_{n\to\infty}\lim\limits_{x\to0}(\frac{1}{n}).(\frac{x}{n})[\large\frac{((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... + 1 ) + ( (1+x/n)^{n-3} + ... + 1 ) + ... + 1)}{x}] =\\$
$=\lim\limits_{n\to\infty}\lim\limits_{x\to 0}\left(\frac{1}{n^2}\right).((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... + 1 ) + ( (1+x/n)^{n-3} + ... + 1 ) + ... + 1) =$
$=\lim\limits_{n\to\infty}\left(\frac{1}{n^2}\right)(n-1 + n-2 + n-3 + ... + 1) = \lim\limits_{n\to\infty}\left(\frac{1}{n^2}\right)(n-1)(\frac{n}{2}) = \lim\limits_{n\to\infty}\frac{n-1}{2n} = \boxed{\large\frac{1}{2}}$.
A: $$
\displaylines{
  \mathop {\lim }\limits_{_{x \to 0} } \frac{{e^x  - x - 1}}{{x^2 }} = \frac{1}{4}\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^{2t}  - 2t - 1}}{{t^2 }} \cr 
   = \frac{1}{4}\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^{2t}  - 2e^t  + 1 - 1 - 2t - 1}}{{t^2 }} \cr 
   = \frac{1}{4}\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^{2t}  - 2e^t  + 1}}{{t^2 }} - 2\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^t  - t - 1}}{{t^2 }} \cr 
  \mathop {\lim }\limits_{_{x \to 0} } \frac{{e^x  - x - 1}}{{x^2 }} = \frac{1}{2} \cdots \left( 1 \right) \cr}
$$
$$
m = \frac{1}{4}\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^{2t}  - 2e^t  + 1}}{{t^2 }} - 2m \Leftrightarrow m = \frac{1}{2}
$$
