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).

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Why don't you want to use them ? – Belgi Aug 18 '12 at 17:00
This amounts to finding the second derivative of $e^x$ at $x=0$. So I guess it's important to motivate why you want to restrict methods of proof. Certainly an approach along the lines that being its own derivative characterizes $e^x$ seems viable. – hardmath Aug 18 '12 at 17:03
What is your definition of $e^x$? – Chris Eagle Aug 18 '12 at 17:05
@enzotib: You could do any number of things. Are you doing that? If yes, what is your definition of $e$? – Chris Eagle Aug 18 '12 at 17:14
@JoelCohen But that would encourage people not to use l'Hopital's, which is a foolish handicap to impose on oneself. – Alex Becker Aug 18 '12 at 17:15

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*}

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very good solution. I think a student who doesnt have knowledge about l'Hopital's rule or any series expansion can simply follow your way))))) – Seyhmus Güngören Aug 18 '12 at 19:04
+1 For the stamina shown in the elegant, though long, solution. Yet, I think many students who haven't yet any knowledge of derivatives, L'H rule and series expansion won't probably have the capability to fully understand double limits and ths swift use of summatories. – DonAntonio Aug 19 '12 at 2:31
Nice way to compute the limit! (+1) – user 1618033 Aug 25 '12 at 15:21
very elegant solution. This goes on to say what is possible to achieve without a direct use of LHR. Personally I try to avoid to LHR if a limit problem can be solved using rules of algebra of limits. But in this case we have to resort to infinite series and double limits (although of an easy kind), so that LHR is much better suited to this. – Paramanand Singh Jan 13 '14 at 3:34

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).

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"if at all possible" - ofcourse it is possible, the answer above showed what it is exactly – Belgi Aug 19 '12 at 10:56
@Belgi: of course, but I mean proving that (without calculating the resulting value) in a simpler way. – enzotib Aug 19 '12 at 11:00
@Belgi: No, there is a subtlety here. For example, consider the series $\sum_{i=0}^{\infty} (-1)^i = 1 - 1 + 1 - 1 + 1 \cdots$. If the sum has a limit, $L$, then it is clear that $L = 1 + \sum_{i=1}^{\infty} (-1)^i = 1 - \sum_{i=0}^{\infty} (-1)^i$. Therefore, since $L = 1 - L$, $L = 1/2$. However, since the limit of partial sums $\sum_{i=1}^{N} (-1)^i$ does not exist, it cannot be equal to $1/2$. – Shaun Ault Aug 19 '12 at 12:35
@ShaunAult - but considering the above answer the limit of the partial sums does exist so it is possible to show that $l$ exist and is finite, this is the only thing I said (I did not say this in general for other sums) – Belgi Aug 19 '12 at 13:03
This answer is really interesting. (+1) – user 1618033 Aug 25 '12 at 15:21

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,

$L = \lim\limits_{x\to 0}\frac{P-1}{x} =\lim\limits_{x\to 0}\lim\limits_{n\to\infty}(\frac{1}{n}).(\frac{x}{n})[\large\frac{((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... + 1 ) + ( (1+x/n)^{n-3} + ... + 1 ) + ... + 1)}{x}]=$

$=\lim\limits_{n\to\infty}\lim\limits_{x\to0}(\frac{1}{n}).(\frac{x}{n})[\large\frac{((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... + 1 ) + ( (1+x/n)^{n-3} + ... + 1 ) + ... + 1)}{x}] =\\$

$=\lim\limits_{n\to\infty}\lim\limits_{x\to 0}\left(\frac{1}{n^2}\right).((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... + 1 ) + ( (1+x/n)^{n-3} + ... + 1 ) + ... + 1) =$

$=\lim\limits_{n\to\infty}\left(\frac{1}{n^2}\right)(n-1 + n-2 + n-3 + ... + 1) = \lim\limits_{n\to\infty}\left(\frac{1}{n^2}\right)(n-1)(\frac{n}{2}) = \lim\limits_{n\to\infty}\frac{n-1}{2n} = \boxed{\large\frac{1}{2}}$.

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$$\displaylines{ \mathop {\lim }\limits_{_{x \to 0} } \frac{{e^x - x - 1}}{{x^2 }} = \frac{1}{4}\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^{2t} - 2t - 1}}{{t^2 }} \cr = \frac{1}{4}\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^{2t} - 2e^t + 1 - 1 - 2t - 1}}{{t^2 }} \cr = \frac{1}{4}\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^{2t} - 2e^t + 1}}{{t^2 }} - 2\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^t - t - 1}}{{t^2 }} \cr \mathop {\lim }\limits_{_{x \to 0} } \frac{{e^x - x - 1}}{{x^2 }} = \frac{1}{2} \cdots \left( 1 \right) \cr}$$

$$m = \frac{1}{4}\mathop {\lim }\limits_{_{t \to 0} } \frac{{e^{2t} - 2e^t + 1}}{{t^2 }} - 2m \Leftrightarrow m = \frac{1}{2}$$

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Why should the limit exist? – egreg Jan 5 '15 at 12:01
Why the additional $-2e^t$ on the second line? – enzotib Jan 6 '15 at 12:03