How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion?

by enzotib   Last Updated March 11, 2017 11:20 AM

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).



Answers 6


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*}

Julian Rosen
Julian Rosen
August 18, 2012 17:43 PM

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).

enzotib
enzotib
August 19, 2012 10:36 AM

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,

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

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

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

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

miguel747
miguel747
January 13, 2014 03:22 AM

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

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

sabachir
sabachir
January 05, 2015 11:58 AM

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!

Dr. MV
Dr. MV
November 14, 2016 22:14 PM

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 $n\rightarrow \infty$.

H. H. Rugh
H. H. Rugh
March 11, 2017 11:00 AM

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