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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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    $\begingroup$ Why don't you want to use them ? $\endgroup$ – Belgi Aug 18 '12 at 17:00
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    $\begingroup$ 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. $\endgroup$ – hardmath Aug 18 '12 at 17:03
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    $\begingroup$ What is your definition of $e^x$? $\endgroup$ – Chris Eagle Aug 18 '12 at 17:05
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    $\begingroup$ @ChrisEagle: I could take $e^x$ as the $\sup$ of $A=\{e^q|q\in\mathbb{Q},q\leq x\}$. $\endgroup$ – enzotib Aug 18 '12 at 17:11
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    $\begingroup$ @enzotib: You could do any number of things. Are you doing that? If yes, what is your definition of $e$? $\endgroup$ – Chris Eagle Aug 18 '12 at 17:14
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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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    $\begingroup$ 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))))) $\endgroup$ – Seyhmus Güngören Aug 18 '12 at 19:04
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    $\begingroup$ +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. $\endgroup$ – DonAntonio Aug 19 '12 at 2:31
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    $\begingroup$ Nice way to compute the limit! (+1) $\endgroup$ – user 1357113 Aug 25 '12 at 15:21
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    $\begingroup$ 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. $\endgroup$ – Paramanand Singh Jan 13 '14 at 3:34
  • $\begingroup$ Julian, nice solution. Just curious as to the relevance of the passage regarding $f(x)$ bounds in terms the supremum and infimum of $e^{p/q}$. In the development, it was assumed that $x=p/q$; the result was that $e^{p/q}=\lim_{n\to \infty}\left(1+\frac{p/q}{n}\right)^n$. Are you trying to establish that the limit extends from the rational numbers to the reals? $\endgroup$ – Mark Viola Nov 14 '16 at 22:28
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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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  • $\begingroup$ "if at all possible" - ofcourse it is possible, the answer above showed what it is exactly $\endgroup$ – Belgi Aug 19 '12 at 10:56
  • $\begingroup$ @Belgi: of course, but I mean proving that (without calculating the resulting value) in a simpler way. $\endgroup$ – enzotib Aug 19 '12 at 11:00
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    $\begingroup$ @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 $1/2$. $\endgroup$ – Shaun Ault Aug 19 '12 at 12:35
  • $\begingroup$ @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) $\endgroup$ – Belgi Aug 19 '12 at 13:03
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    $\begingroup$ This answer is really interesting. (+1) $\endgroup$ – user 1357113 Aug 25 '12 at 15:21
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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!

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

$$ e^x-1= \sum_{k=0}^{n-1} e^{\frac{k}{n} x} \left( e^{\frac{x}{n}} -1\right)= \sum_{k=0}^{n-1} \left( 1+ \frac{k}{n}x + R(\frac{k}{n}x) \right) \left( \frac{x}{n} + R(\frac{x}{n}) \right) $$ Developping 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: Incidently 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 $n\rightarrow \infty$.

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