Computing $\lim_{x\rightarrow 0}{\frac{xe^x- e^x + 1}{x(e^x-1)}}$ without L'Hôpital's rule or Taylor series

Question

This limit really stamped me because i'm not allowed to use L'Hôpital's rule or Taylor's series, please help!

I think the limit is $\frac{1}{2}$, but i don't know how to prove it without the L'Hôpital's rule or Taylor's series

$$\lim_{x\rightarrow 0}{\frac{xe^x- e^x + 1}{x(e^x-1)}}$$

Answer 1

Having $$\lim\limits_{x\rightarrow 0}{\frac{xe^x- e^x + 1}{x(e^x-1)}}= 1+\lim\limits_{x\rightarrow 0}{\frac{x- e^x + 1}{x(e^x-1)}}= 1+\lim\limits_{x\rightarrow 0}{\frac{x- e^x + 1}{x^2\frac{e^x-1}{x}}}= 1-\lim\limits_{x\rightarrow 0}{\frac{e^x-1 - x}{x^2}}$$ It's only left to compute $\lim\limits_{x\rightarrow 0}{\frac{e^x-1 - x}{x^2}}$, which is not that trivial seing the answers to that question.

Answer 2

Replacing $ x $ by $\color{red}{ -x} $,

$$L=\lim_0\frac{xe^x-e^x+1}{x(e^x-1)}$$ $$=\lim_0\frac{-xe^{\color{red}{-x}}-e^{-x}+1}{-x(e^{-x}-1)}$$

$$=\lim_0\frac{-x-1+e^x}{x(e^x-1)}$$

the sum gives $$2L=\lim_0\frac{x(e^x-1)}{x(e^x-1)}=1$$ thus $$L=\frac 12$$

Answer 3

How about using the Cauchy's mean value theorem (L'Hospital rule can be seen as a specialization of this). Let $f(x)=xe^x-e^x+1$ and $g(x)=xe^x-x$, then $f(0)=0=g(0)$ and by the (generalize) mean value theorem, there is $c_x$ between $0$ and $x$ such that $$f'(c_x)(g(x)-g(0))=g'(c_x)(f(x)-f(0)).$$ This can be expressed as

$$ \frac{f(x)}{g(x)}=\frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(c_x)}{g'(c_x)}=\frac{c_xe^{c_x}}{c_xe^{c_x}+ e^{c_x}-1}=\frac{e^{c_x}}{e^{c_x}+\frac{e^{c_x}-1}{c_x}}$$

As $x\rightarrow 0$, $c_x\rightarrow 0$ and so

$$\lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow0}\frac{e^{c_x}}{e^{c_x}+\frac{e^{c_x}-1}{c_x}}=\frac{1}{2}$$

Here we have use the fact that $\lim_{h\rightarrow0}\frac{e^h-1}{h}=\exp'(0)=1$.

Answer 4

First of all, let us compute $ \lim\limits_{x\to 0}{\frac{\mathrm{e}^{-x}+x-1}{x^{2}}} $:

Notice that for any $ t\in\mathbb{R} $, $ \left|\mathrm{e}^{t}-1\right|=\left|t\right|\left|\int_{0}^{1}{\mathrm{e}^{xt}\,\mathrm{d}x}\right|\leq\left|t\right|\int_{0}^{1}{\mathrm{e}^{x\left|t\right|}\,\mathrm{d}x}\leq\left|t\right|\mathrm{e}^{\left|t\right|} \cdot $

Observe that : \begin{aligned} \frac{\mathrm{e}^{-x}+x-1}{x^{2}}&=\int_{0}^{1}{\left(1-y\right)\mathrm{e}^{-xy}\,\mathrm{d}y}\\ &=\frac{1}{2}+\int_{0}^{1}{\left(1-y\right)\left(\mathrm{e}^{-xy}-1\right)\mathrm{d}y} \end{aligned}

Since $ \left|\int_{0}^{1}{\left(1-y\right)\left(\mathrm{e}^{-xy}-1\right)\mathrm{d}y}\right|\leq\int_{0}^{1}{\left(1-y\right)\left|\mathrm{e}^{-xy}-1\right|\mathrm{d}y}\leq \left|x\right|\int_{0}^{1}{y\left(1-y\right)\mathrm{e}^{\left|x\right|y}\,\mathrm{d}y}\underset{x\to 0}{\longrightarrow}0 $, we get :

$$ \frac{\mathrm{e}^{-x}+x-1}{x^{2}}\underset{x\to 0}{\longrightarrow}\frac{1}{2} $$

And thus : \begin{aligned}\lim_{x\to 0}{\frac{x\,\mathrm{e}^{x}-\mathrm{e}^{x}+1}{x\left(\mathrm{e}^{x}-1\right)}}&=\lim_{x\to 0}{\left(\frac{\mathrm{e}^{-x}+x-1}{x^{2}}\times\frac{x}{1-\mathrm{e}^{-x}}\right)}\\ &=\frac{1}{2}\times 1\\ \lim_{x\to 0}{\frac{x\,\mathrm{e}^{x}-\mathrm{e}^{x}+1}{x\left(\mathrm{e}^{x}-1\right)}}&=\frac{1}{2}\end{aligned}

Answer 5

We will use the definition $e^x:=\lim_{n \to \infty} \left( 1+ \frac{x}{n}\right)^n$ which we know to converge for any $x$.

1. Let's first multiply both numerator and denominator by $e^{-x/2}$, which gives \begin{eqnarray} \frac{x e^x - e^x +1}{x(e^x-1)}&=&\frac{x e^{x/2}-e^{x/2}+e^{-x/2}}{x(e^{x/2}-e^{-x/2})}. \end{eqnarray}

2. Let's further note that $e^{x/2} = (e^{x/2}-e^{-x/2})/2 + (e^{x/2}+e^{-x/2})/2$, which gives \begin{eqnarray} \frac{x e^x - e^x +1}{x(e^x-1)}&=&\frac{1}{2}+{\bf{ \frac{1}{2}\frac{(e^{x/2}+e^{-x/2})}{(e^{x/2}-e^{-x/2})} - \frac{1}{x}}}\\ &=&\frac{1}{2}+{\bf{\frac{1}{2} \frac{(x/2)(e^{x/2}+e^{-x/2})-(e^{x/2}-e^{-x/2})}{(x/2)(e^{x/2}-e^{-x/2})} }}. \end{eqnarray}

   Therefore, the desired limit is \begin{eqnarray} L=\lim_{x \to 0}\left[\frac{x e^x - e^x +1}{x(e^x-1)}\right]&=&\frac{1}{2}+\lim_{x \to 0} { \bf{\left[\frac{1}{2} Z(x/2)\right]}}, \end{eqnarray} where $Z(x)={ \frac{(e^{x}+e^{-x})}{(e^{x}-e^{-x})} - \frac{1}{x}}={ \frac{x (e^{x}+e^{-x})-(e^{x}-e^{-x})}{x(e^{x}-e^{-x})} }$.

   Let's now prove that $Z=\lim_{x \to 0}{\left[Z(x)\right]}$ exists and is zero, and, therefore, that $L=1/2$.

3. Let's restrict ourselves to even terms, $n=2 p$, in the definition of $e^x$ -we can do that, as it converges-, i.e. \begin{eqnarray} e^x = \lim_{n \to \infty} \left( 1+ \frac{x}{n}\right)^n = \lim_{p \to \infty} \left( 1+ \frac{x}{2 p}\right)^{2p} \end{eqnarray} and introduce this in the expression of $Z(x)$.

   The sum and difference of positive and negative exponential functions can thus be written as \begin{eqnarray} \left(e^x \pm e^{-x} \right) = \lim_{p \to \infty} \left[\left( 1+ \frac{x}{2 p}\right)^{2p} \pm \left( 1 - \frac{x}{2 p}\right)^{2p} \right]. \end{eqnarray} Using the binomial expansion $(1 + b)^{2p} = \sum_{k=0}^{2 p} \left[\frac{(2 p)!}{(2 p -k)! k!} b^k\right]$, for $b=x/(2 p)$ or $-x/(2 p)$, we find that only the even powers of $x$ remain in the sums and only the odd powers in the differences, so that $ \left(e^x + e^{-x} \right) = \lim_{p \to \infty}\left[ P_p(x) \right] $ and $ \left(e^x - e^{-x} \right) = \lim_{p \to \infty}\left[ M_p(x) \right]$, where
   \begin{eqnarray} P_p(x)&=&2 \sum_{k=0}^p \left[ \frac{(2 p)!}{(2 p- 2 k)! (2 k)!} \frac{x^{2 k}}{(2 p)^{2 k}}\right] \\ M_p(x)&=&2 \sum_{k=1}^p \left[ \frac{(2 p)!}{(2 p- 2 k +1)! (2 k - 1)!} \frac{x^{2 k -1}}{(2 p)^{2 k -1}}\right]=2 \sum_{k'=0}^{p -1} \left[ \frac{(2 p)!}{(2 p- 2 k' -1)! (2 k' + 1)!} \frac{x^{2 k' +1}}{(2 p)^{2 k' +1}}\right], \end{eqnarray} and \begin{eqnarray} Z=\lim_{x \to 0}{\left[Z(x)\right]}=\lim_{x \to 0}\left\{ \frac{\lim_{p \to \infty}\left[x P_p(x)\right]- \lim_{p \to \infty} M_p(x)}{ \lim_{p \to \infty}\left[x M_p(x) \right]}\right\}. \end{eqnarray}

4. Combining and comparing the terms in the three converging series, $\lim_{p \to \infty}\left[x P_p(x)\right]$, $\lim_{p \to \infty}\left[M_p(x)\right]$, and $\lim_{p \to \infty}\left[x M_p(x)\right]$, we can now show that \begin{eqnarray} Z(x)=x .\frac{ f(x)}{g(x)} \end{eqnarray} for some well defined functions $f(x)$ and $g(x)$ which further satisfy $0<f(x)<g(x)$ for any $x\neq 0$. This ensures that \begin{eqnarray} Z = \lim_{x \to 0}{Z(x)}=\lim_{x \to 0}{\frac{x. f(x)}{g(x)}}=0. \end{eqnarray}

5. Coming back to the expression of $Z(x)$, at the end of point 3, and substituting the series, \begin{eqnarray} &Z(x)=\frac{\lim_{p \to \infty}\left[x P_p(x)\right]- \lim_{p \to \infty} M_p(x)}{ \lim_{p \to \infty}\left[x M_p(x) \right]}&\\ &=\frac{\lim_{p \to \infty}\left\{ \sum_{k=0}^p \left[ \frac{(2 p)! (2 p)}{(2 p- 2 k)! (2 k)!} \frac{x^{2 k +1}}{(2 p)^{2 k +1}}\right]\right\}- \lim_{p \to \infty} \left\{\sum_{k'=0}^{p -1} \left[ \frac{(2 p)!}{(2 p- 2 k' -1)! (2 k' + 1)!} \frac{x^{2 k' +1}}{(2 p)^{2 k' +1}}\right]\right\}}{ \lim_{p \to \infty}\left\{\sum_{k=1}^p \left[ \frac{(2 p)! (2 p)}{(2 p- 2 k +1)! (2 k - 1)!} \frac{x^{2 k }}{(2 p)^{2 k }}\right] \right\}}.& \end{eqnarray} One can group the individual terms of the converging series at the numerator and factorise $x$, that gives \begin{eqnarray} Z(x)=\frac{x \lim_{p \to \infty}\left( \sum_{k=0}^{p-1} \left\{ \left[\frac{(2 p)! (2 p)}{(2 p- 2 k)! (2 k)!}- \frac{(2 p)!}{(2 p- 2 k -1)! (2 k + 1)!} \right] \frac{x^{2 k}}{(2 p)^{2 k +1}} \right\} + \frac{x^{2 p}}{(2 p)^{2 p}} \right)} { \lim_{p \to \infty}\left\{\sum_{k=1}^{p-1} \left[ \frac{(2 p)! (2 p)}{(2 p- 2 k +1)! (2 k - 1)!} \frac{x^{2 k }}{(2 p)^{2 k }}\right] + (2 p)^2 \frac{x^{2 p }}{(2 p)^{2 p }} \right\}}. \end{eqnarray} (Note that, both at the numerator and the denominator, the last term is added after the sum of all the others terms.)

   Three things are to be noted to complete the proof:

   • the first term of the sum at the numerator is zero, since $\left[\frac{(2 p)! (2 p)}{(2 p- 2 k)! (2 k)!}- \frac{(2 p)!}{(2 p- 2 k -1)! (2 k + 1)!} \right] =0$ for $k=0$;
   • all terms of the sums are positive: all powers of $x$ are even and therefore are positive, even if $x<0$, and for all coefficients ($k=1, \dots, p-1$) at the numerator, one has \begin{eqnarray} (2 p)! \frac{ (2 p)}{(2 p- 2 k)! (2 k)!} > (2 p)! \frac{1}{(2 p- 2 k -1)! (2 k + 1)!} = (2 p)! \frac{(2 p- 2 k)}{(2 p- 2 k)! (2 k)!}\frac{1}{(2 k + 1)}, \end{eqnarray} since $(2 p)>{(2 p- 2 k)}/{(2 k + 1)}$;
   • the coefficients of the powers of x at the numerator are all smaller than the corresponding coefficients at the denominator, since for $k=1, \dots, p-1$, $1/(2 k)<2 p/(2 p - 2 k +1) = 1 + (2 k -1)/(2 p - 2 k +1)$, and since, for the '$x^{(2 p)}$' terms, $1 < (2 p)^2$.

That's it.

Note

Note the incidental proof that the function $Z(x)/x < c=1$, or \begin{eqnarray} { \frac{x (e^{x}+e^{-x})-(e^{x}-e^{-x})}{x^2(e^{x}-e^{-x})} } &<& c \\ (-1 + x. \coth(x)) &<& c. x^2 \\ \dots \end{eqnarray} A tighter bound would be $c=1/3$ that is reached at $x=0$, i.e. $Z(x)/x < 1/3$ for $x \neq 0$ and $\lim_{x \to 0} Z(x)/x = 1/3$.