Is it possible to determine the limit


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
    Commented Aug 18, 2012 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
    Commented Aug 18, 2012 at 17:03
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    $\begingroup$ What is your definition of $e^x$? $\endgroup$ Commented Aug 18, 2012 at 17:05
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    $\begingroup$ @ChrisEagle: I could take $e^x$ as the $\sup$ of $A=\{e^q|q\in\mathbb{Q},q\leq x\}$. $\endgroup$ Commented Aug 18, 2012 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 $e$? $\endgroup$ Commented Aug 18, 2012 at 17:14

7 Answers 7


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))))) $\endgroup$ Commented Aug 18, 2012 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
    Commented Aug 19, 2012 at 2:31
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    $\begingroup$ Nice way to compute the limit! (+1) $\endgroup$ Commented Aug 25, 2012 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
    Commented Jan 13, 2014 at 3:34
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    $\begingroup$ 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
    Commented Nov 14, 2016 at 22:28

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).

  • $\begingroup$ "if at all possible" - ofcourse it is possible, the answer above showed what it is exactly $\endgroup$
    – Belgi
    Commented Aug 19, 2012 at 10:56
  • $\begingroup$ @Belgi: of course, but I mean proving that (without calculating the resulting value) in a simpler way. $\endgroup$ Commented Aug 19, 2012 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
    Commented Aug 19, 2012 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
    Commented Aug 19, 2012 at 13:03
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    $\begingroup$ This answer is really interesting. (+1) $\endgroup$ Commented Aug 25, 2012 at 15:21

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!


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 $\int_0^xe^u\,du=e^x-1$.

  • 1
    $\begingroup$ Approved your very nice solution. $\endgroup$
    – Sebastiano
    Commented Jul 10, 2020 at 12:05

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 $n\rightarrow \infty$.


Consider fundamental limit: $e = \lim\limits_{n\to \infty}(1+\frac{1}{n})^n$ and $e^x = \lim\limits_{n\to\infty}(1+\frac{x}{n})^n$


$e^x = [\lim\limits_{k\to\infty}(1+1/k)^k]^x = \lim\limits_{k\to \infty}((1+1/k)^{kx})\Rightarrow kx = n \Rightarrow e^x = \lim\limits_{n\to\infty}(1+\frac{x}{n})^n$.

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:

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


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

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

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

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


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

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

  • 2
    $\begingroup$ Why should the limit exist? $\endgroup$
    – egreg
    Commented Jan 5, 2015 at 12:01
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    $\begingroup$ Why the additional $-2e^t$ on the second line? $\endgroup$ Commented Jan 6, 2015 at 12:03

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