I would like to solve that limit solved using only rules of algebra of limits.


All the answers in How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion? do not fully address my question.

A challenging limit problem for the level of student who knows that: $$\begin{align*} \lim\limits_{x\to +\infty} e^x&=+\infty\tag1\\ \lim\limits_{x\to -\infty} e^x&=0\tag2\\ \lim\limits_{x\to +\infty} \frac{e^x}{x^n}&=+\infty\tag3\\ \lim\limits_{x\to -\infty} x^ne^x&=0\tag4\\ \lim\limits_{x\to 0} \frac{e^x-1}{x}&=1\tag5 \end{align*}$$

  • $\begingroup$ you need $\epsilon-\delta$ stuff? $\endgroup$
    – Math-fun
    Mar 7, 2015 at 11:51
  • $\begingroup$ no because the level of student doesn't support that kind of stuff $\endgroup$
    – Educ
    Mar 7, 2015 at 12:01
  • $\begingroup$ Why does the last answer (the 25 + yes answer) not address your question? What about it does not help/work for you? $\endgroup$
    – Alan
    Mar 7, 2015 at 12:04
  • 7
    $\begingroup$ We have to know what definition of $e^x$ to use... $\endgroup$
    – Ian
    Mar 7, 2015 at 12:21
  • 1
    $\begingroup$ Some answers below use the fact that the desired limit exists. Showing that the limit exists is tricky and not possible without the use of derivative (i.e. L'Hospital, Taylor or something equivalent). It is better to understand the powers of rules of algebra of limits but at the same time it is important to know their limitation. $\endgroup$
    – Paramanand Singh
    Mar 10, 2015 at 4:12

8 Answers 8


$$ \lim\limits_{x\to 0}\dfrac{e^x-1-x}{x^2} = \lim\limits_{x\to 0}\left(\dfrac{1}{e^x+1+x}\right)\lim\limits_{x\to 0}\dfrac{e^{2x}-(1+x)^2}{x^2} = \dfrac{1}{2}\lim\limits_{x\to 0}\left(4\dfrac{e^{2x}-1-2x}{4x^2}-1\right) = 2\lim\limits_{x\to 0}\left(\dfrac{e^{2x}-1-2x}{4x^2}\right)-\dfrac{1}{2} = 2\lim\limits_{x\to 0}\left(\dfrac{e^x-1-x}{x^2}\right)-\dfrac{1}{2}. $$


$$\lim\limits_{x\to 0}\dfrac{e^x-1-x}{x^2} = \dfrac{1}{2}\,.$$

This can be extended in a natural way. Define $s_{n}(x):=\sum\limits_{k=0}^{n-1}\dfrac{x^k}{k!}$ and observe the following: $$ \begin{eqnarray*} \lim_{x\to0}\dfrac{e^x-s_{n}(x)}{x^n} &=& \lim_{x\to0}\left(\dfrac{1}{\sum\limits_{r=0}^{n-1}e^{rx}(s_n(x))^{n-1-r}}\right)\lim_{x\to0}\dfrac{e^{nx}-(s_n(x))^n}{x^n} \\ &&\ \\ \ \\ &=& \dfrac{1}{n}\left(\lim_{x\to0}\dfrac{e^{nx}-s_n(nx)-\frac{n^{n-1}-1}{(n-1)!}x^n-o(x^n)}{x^n}\right)\\ &&\ \\ \ \\&=& n^{n-1}\lim_{x\to0}\left(\dfrac{e^{nx}-s_n(nx)}{(nx)^n}\right)-\frac{n^{n-1}-1}{n!}\,, \end{eqnarray*} $$

where, upon expansion, the polynomial $(s_n(x))^n = s_n(nx) + \frac{n^{n-1}-1}{(n-1)!}x^n + o(x^n)$. Consequently,

$$ \lim_{x\to0}\dfrac{e^x-s_n(x)}{x^n} = \dfrac{1}{n!}\,. $$

The foregoing arguments require the existence of the primary limits being sought after. Given that our hypothetical pupil is not armed with sophisticated $\epsilon, \delta$ analysis techniques, it is difficult to see how he might proceed in showing existence. Indeed, how does this pupil even know what a limit is, let alone the algebraic laws that follow by definition? Nevertheless, with these concerns in mind, we might at least allow our student one more atom of information: the interchange of limits for the particular class of functions we are interested in here. Thus, our pupil proceeds, somewhat mechanistically, as follows:

$$ \begin{eqnarray*} \lim\limits_{x\to 0}\frac{e^{x}-s_n(x)}{x^n} &=& \lim\limits_{x\to 0}\lim\limits_{y\to 0}\frac{\left(e^{y}(\frac{e^{x}-1}{x})-\frac{1}{x}\Big(s_n(x+y)-s_n(y)\Big)\right)}{\Big(\sum\limits_{r=0}^{n-1}(x+y)^r y^{n-1-r}\Big)} \\ &&\ \\&=& \lim\limits_{y\to 0}\lim\limits_{x\to 0}\frac{\left(e^{y}(\frac{e^{x}-1}{x})-\frac{1}{x}\Big(s_n(x+y)-s_n(y)\Big)\right)}{\Big(\sum\limits_{r=0}^{n-1}(x+y)^r y^{n-1-r}\Big)} \\ &&\ \\&=& \frac{1}{n}\lim\limits_{y\to 0}\frac{\left(e^{y}-s_{n-1}(y)\right)}{y^{n-1}}\,. \end{eqnarray*}$$

So, by successive application of such interchanges, our pupil deduces

$$ \lim\limits_{x\to 0}\frac{e^{x}-s_n(x)}{x^n} = \frac{1}{n!}\,. $$

  • 3
    $\begingroup$ This is very interesting. Can all the terms of the Taylor series of $e^x$ be done similarly? $\endgroup$
    – GEdgar
    Mar 9, 2015 at 14:01
  • 1
    $\begingroup$ I believe that you use here assumption that the limit exists. Can you show the existence of limit? $\endgroup$
    – Wojowu
    Mar 9, 2015 at 15:20
  • $\begingroup$ @GEdgar, please see edits. $\endgroup$
    – ki3i
    Mar 10, 2015 at 19:20
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    $\begingroup$ nice work. the conjugates. need to remember that. $\endgroup$
    – abel
    Mar 10, 2015 at 23:13
  • 1
    $\begingroup$ Nice trick. Alas it only works if you know the limit exists. $\endgroup$
    – Git Gud
    Mar 10, 2017 at 23:00


Then, taking the limit,


UPDATE: the same approach can be used for the next order,

$$\left(\frac{e^x-1}x\right)^3=\frac{e^{3x}-3e^{2x}+3e^x-1}{x^3}=3^3f(3x)-3\cdot2^3(2x)+3f(x),$$ where $f(x)=\dfrac{e^x-1-x-\dfrac{x^2}2}{x^3},$ and $$1=27L-24L+3L.$$

  • $\begingroup$ Just a shorter version than that of @ki3i. $\endgroup$
    – user65203
    Mar 9, 2015 at 13:47
  • 1
    $\begingroup$ I originally thought the question was kind of stupid. But I see from these two answers that I was wrong. $\endgroup$
    – GEdgar
    Mar 10, 2015 at 21:43
  • 1
    $\begingroup$ @GEdgar: I was myself amazed and pleased to discover that the whole Taylor development can be derived from the initial limit. $\endgroup$
    – user65203
    Mar 11, 2015 at 7:35

In order to show that

$$\lim_{x\to0}{e^x-1-x\over x^2}={1\over2}$$

it suffices to show that

$$\lim_{x\to0}{1\over x^2}\int_0^x(e^u-1-u)\,du=0$$

Let $g(u)=e^u-1-u$. Since $g'(u)=e^u-1$ and $g''(u)=e^u$, we see that $g(u)$ has a global minimum at $u=0$, with $g(0)=0$. Consequently

$$0\le{1\over x^2}\int_0^x(e^u-1-u)\,du\le{1\over x}\int_0^x{e^u-1-u\over u}\,du\quad\text{for }x\not=0$$

By the Mean Value Theorem,

$$\lim_{x\to0}{1\over x}\int_0^x{e^u-1-u\over u}\,du=\lim_{u\to0}{e^u-1-u\over u}$$


$$\lim_{u\to0}{e^u-1-u\over u}=\lim_{u\to0}{g(u)-g(0)\over u-0}=g'(0)=0$$

Thus, by the Squeeze Theorem, $\lim_{x\to0}{1\over x^2}\int_0^x(e^u-1-u)\,du=0$.


Note that

$$e^x-1-x= \int_0^x (e^t-1)\, dt = \int_0^x \int_0^t e^s\,ds\, dt.$$

Now $1\le e^s \le e^x$ above. Thus $e^x-1-x$ lies between

$$\tag 1 \int_0^x \int_0^t 1 \,ds\, dt,\, \,\,\int_0^x \int_0^t e^x\,ds\, dt.$$

The first iterated integral in $(1)$ equals $x^2/2,$ the one on the right equals $e^x(x^2/2).$ So we have $e^x-1-x$ between $x^2/2$ and $e^x(x^2/2).$ Dividing by $x^2$ and using the squeeze theorem gives $1/2$ for the desired limit.

  • $\begingroup$ I managed to extend this method to the more general case, if you'd care to take a look. $\endgroup$ Aug 7, 2017 at 12:06
  • $\begingroup$ (+1) for efficiency! Happy Holidays my friend. Hope you're doing well! $\endgroup$
    – Mark Viola
    Nov 26, 2019 at 20:18
  • $\begingroup$ @MarkViola MV, I haven't seen you in the longest! May the blessings of the season be with you. $\endgroup$
    – zhw.
    Nov 26, 2019 at 22:18
  • $\begingroup$ @zhw. Thank you ZW! And you. $\endgroup$
    – Mark Viola
    Nov 26, 2019 at 22:26

One could consider the more general limit:




It may be easily seen that

$$e^x-P_n(x)=\int_0^xe^t-P_{n-1}(t)~\mathrm dt$$

And by induction,

$$e^x-P_{n-1}(x)=\int_0^x\int_0^{\sigma_n}\dots\int_0^{\sigma_2}e^{\sigma_1}~\mathrm d\sigma_1~\mathrm d\sigma_2\dots\mathrm d\sigma_n$$

By Cauchy's repeated integral formula, it can be seen that

$$e^x-P_{n-1}(x)=\frac1{(n-1)!}\int_0^x(x-t)^{n-1}e^t~\mathrm dt$$

And by substituting $t=xu$, this simplifies down to

$$e^x-P_{n-1}(x)=\frac{x^n}{(n-1)!}\int_0^1(1-u)^{n-1}e^{xu}~\mathrm du$$

Which so nicely cancels the $x^n$, and likewise, it is easy to see that

$$\int_0^1(1-u)^{n-1}e^{-|x|}~\mathrm du<\int_0^1(1-t)^{n-1}e^{xu}~\mathrm du<\int_0^1(1-u)^{n-1}e^{|x|}~\mathrm du$$

So by the squeeze theorem,

$$\begin{align}\lim_{x\to0}x^{-n}(e^x-P_{n-1}(x))&=\lim_{x\to0}\frac1{(n-1)!}\int_0^1(1-u)^{n-1}e^{xu}~\mathrm du\\&=\frac1{(n-1)!}\int_0^1(1-u)^{n-1}~\mathrm du\\&=\frac1{(n-1)!}\frac1n\\&=\frac1{n!}\end{align}$$

As expected.

  • $\begingroup$ This is effectively the Taylor's theorem with Cauchy's remainder applied on exponential function. +1 $\endgroup$
    – Paramanand Singh
    Aug 7, 2017 at 12:37

$$\lim_{x\to0} \dfrac{e^x-1-x}{x^2}= \lim_{x\to0} \dfrac{e^{-x}-1+x}{x^2}= \lim_{x\to0} \dfrac{e^x+e^{-x}-2}{2x^2}=\displaystyle \dfrac{1}{2} \lim_{x\to0} \Bigg(\dfrac{e^{\frac{x}{2}}-e^{\frac{-x}{2}}}{x}\Bigg)^2$$ $$=\displaystyle \dfrac{1}{2} \lim_{x\to0} \Bigg(\dfrac{e^{\frac{x}{2}}-1}{x}-\dfrac{e^{\frac{-x}{2}}-1}{x}\Bigg)^2=\dfrac{1}{2} \Bigg(\dfrac{1}{2}-\dfrac{-1}{2}\Bigg)^2=\dfrac{1}{2}$$


Substitute $h=x^2$: it's $$\lim_{h \to 0} \frac{e^{\sqrt{h}}-\sqrt{h}-1}h$$ which is the derivative of $x \mapsto e^{\sqrt{x}}-\sqrt{x}$ evaluated at $x=0$.

Now use the chain rule: it's $$\lim_{x \to 0} \frac{e^{\sqrt{x}}-1}{2 \sqrt{x}} = \frac{1}{2}\lim_{h \to 0} \frac{e^h - 1}{h}$$

That latter limit is $\exp'(0)= 1$.

  • $\begingroup$ You are not trying to apply chain rule but rather trying to implicitly use the fact that if $g(x) =e^{\sqrt{x}} - \sqrt{x} $ then $g'$ is continuous on $[0,\infty)$. This requires that you already know that $g'(0)=1$. Instead you have proved that $\lim_{x\to 0}g'(x)=1$. $\endgroup$
    – Paramanand Singh
    Oct 8, 2018 at 14:25

$$e^x=1+x+\frac{1}{2}x^2+o(x^2)_{x\to0}\\ \lim_{x\to0}\frac{e^x-1-x}{x^2}=\lim_{x\to0}\frac{1+x+\frac{1}{2}x^2+o(x^2)-1-x}{x^2}= \lim_{x\to0}\frac{\frac{1}{2}x^2+o(x^2)}{x^2}=\frac{1}{2}+0 $$ ($f=o(x^2)\Leftrightarrow\lim_{x\to0}\frac{f(x)}{x^2}=0$)

  • 1
    $\begingroup$ i said only rules of algebra of limits and without expansion series $\endgroup$
    – Educ
    Mar 9, 2015 at 13:58

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