# Evaluate $\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$

Somebody asked this and I think it's quite interesting as I couldn't figure out how to evaluate this but the Wolfram Alpha says its limit is $\frac e2$. $$\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$$ Could someone help here?

## 6 Answers

After playing around a bit, it becomes clear that $f(x) = \frac{\log(1+x)}{x}$ is an important function for this problem; we can see that $f(0) = 1$ easily enough, but we are also interested in $f'(0)$. This function is analytic, and its Taylor expansion at $x = 0$ (using the Taylor expansion for $\log(1+x)$) is $$f(x) = \frac{- \sum_{k=1}^{\infty} \frac{(-1)^k x^k}{k}}{x} = -\sum_{k=1}^{\infty} \frac{(-1)^k x^{k-1}}{k} = \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k+1}\text{.}$$ Dropping the constant term and calculating the derivative, we get \begin{align*} f'(x) &= \sum_{k=1}^{\infty}\frac{k (-1)^k x^{k-1}}{k+1} \\ f'(0) &= -\frac{1}{2} \end{align*}

Back to the original problem. \begin{align*} \lim_{x \to 0} \frac{e - (1+x)^{\frac{1}{x}}}{x} &= \lim_{x \to 0} \frac{e - e^{f(x)}}{x} \\ &= \lim_{x \to 0} \frac{-f'(x)e^{f(x)}}{1} \\ &= - f'(0) e^{f(0)} \\ &= \frac{e}{2} \text{.} \end{align*}

• Thanks. I like this answer better as it didn't use l'Hopital (although it may be basically same to the Rene's answer). Let me choose this answer as the one who asked me on this was a high school teacher (and I guess she got the question from her student) so avoiding l'Hopital is good. – Kay K. Sep 11 '16 at 2:29
• @KayK. Note that, as written, I used l'Hospital's rule in the second part. You could rework it without too much trouble, though, I think, observing that the limit is asking for the derivative of another function involving the same $f(x)$, along the lines of what 王李远 suggested, and using my first half to compute it. – Mike Haskel Sep 11 '16 at 2:33
• I accepted the Rene's answer but I'll go along your approach to put together my answer to avoid l'Hopital. Thanks for all your help. – Kay K. Sep 11 '16 at 2:34
• Mike, yes that was what I meant by not using l'Hopital. – Kay K. Sep 11 '16 at 2:36
• Yeah. I was thinking $\frac{e-e^{f(x)}}{x}=\frac{e^0-e^{f(x)}}{f(x)}\cdot\frac{f(x)-f(0)}{x}$. – Kay K. Sep 11 '16 at 2:37

This limit can be evaluated by applying l'Hospital's rule twice. For the first time we differentiate

$$(1+x)^{\frac{1}{x}}=e^{\frac{\ln (1+x)}{x}}$$ to get

$$\frac{\frac{x}{1+x}-\ln (1+x)}{x^2} e^{\frac{\ln (1+x)}{x}}.$$

Now $$e^{\frac{\ln (1+x)}{x}}\rightarrow e$$ so we need the limit of $$\frac{\frac{x}{1+x}-\ln (1+x)}{x^2}$$ another application of l'Hosptial gives

$$\frac{\frac{1}{(1+x)^2}-\frac{1}{1+x}}{2x}=-\frac{1}{2}\frac{1}{(1+x)^2}\rightarrow -\frac{1}{2}$$

So the limit is $-\frac{e}{2}$.

• Damn it, that's what happens when you don't watch the screen while typing out an answer. Yours is cleaner, +1. – Mike Haskel Sep 11 '16 at 2:26
• @MikeHaskel Hey Mike, if you think so then let me choose this one. Sorry about my change :-) – Kay K. Sep 11 '16 at 2:32
• Note that there's a sign error, since the original limit is actually the derivative of $-(1+x)^{\frac{1}{x}}$ (with a minus sign) evaluated at $x = 0$. – Mike Haskel Sep 11 '16 at 2:53

We can proceed as follows \begin{align} L &= \lim_{x \to 0}\frac{e - (1 + x)^{1/x}}{x}\notag\\ &= \lim_{x \to 0}\dfrac{\exp(1) - \exp\left(\dfrac{\log(1 + x)}{x}\right)}{x}\notag\\ &= -e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - 1}{x}\notag\\ &= -e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - 1}{\dfrac{\log(1 + x)}{x} - 1}\cdot\dfrac{\dfrac{\log(1 + x)}{x} - 1}{x}\notag\\ &= -e\lim_{x \to 0}\frac{\log(1 + x) - x}{x^{2}}\notag \end{align} The last limit can be easily evaluated either by Taylor series, or via L'Hospital's Rule or using integration and the value of this limit is $-1/2$ so that the answer to our question is $e/2$.

$g(x) = \dfrac{\ln(1+x)}{x} = 1 - \dfrac{x}{2} + \dfrac{x^2}{3} + \dots$ is analytic on $(-1,1)$ which implies $h(x) = \exp g(x)$ has a power series expansion around $0$ say $$h(x) = a_0 + a_1 x + a_2 x^2 + \dots$$

We have $a_0 = h(0) = \exp g(0) = e$,

$a_1 = h'(0) = g'(0) \exp g(0) = -e/2$

so

$\dfrac{h(x) - e}{x} \to a_1$ as $x \to 0$ i.e., $\lim_{x\to0}\dfrac{(1+x)^{1/x}-e}{x} = -\dfrac{e}{2}$ or $\lim_{x \to 0}\dfrac{e - (1+x)^{1/x}}{x} = \dfrac{e}{2}.$

For $0<x<1$: We have $$\frac {1}{x}\ln (1+x)=1-x/2+(x^2/3-x^3/4)+(x^4/5-x^5/6)+...=$$ $$=1-x/2+x^2/3-(x^3-x^4/5)-(x^5-x^6/7)-...$$ Therefore $$1- x/2<\frac {1}{x}\ln (1+x)<1-x/2+x^2/3.$$ Therefore $$e(1-e^{-x/2+x^2/3})<e-(1+x)^{\frac {1}{x}} <e(1-e^{-x/2}).$$

Both $(-x/2+x^2/3)$ and $(-x/2)$ belong to the interval $(-1,0).$ When $y\in (-1,0)$ we have $$1+y<1+y+(y^2/2!+y^3/3!)+(y^4/4!+y^5/5!)+...=e^y$$ $$=1+y+y^2/2!+(y^3/3!+y^4/4)+..<1+y+y^2/2!.$$ So $$-y_1-y_1^2/2=1-(1+y_1+y_1^2/2!))<1-e^{y_1}$$ where $y_1=-x/2+x^2/3.$..... And also $$1-e^{-x/2}<1-(1-x/2!)=x/2.$$ Now $y_1+y_1^2/2=-x/2 +x^2F(x)$ where $F(x)$ is a polynomial, so for some $K>0$ we have $x\in (0,1)\implies |F(x)|<K.$ Therefore $$e/2-xK<\frac {e-(1+x)^{\frac {1}{x}}}{x}<e/2.$$

You can think of the limit $$\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$$ as the derivative of the function $f(x)=(1+x)^\frac1x$ at point $x=0$.

Did that idea gave you any help?

Edit: You should define the value of $f(x)$ at $x=0$ namely : $f(0)=e$

• I already tried it but $f'(0)$ looked even more difficult to evaluate. – Kay K. Sep 11 '16 at 1:41
• To use that, you need to show that $(1+x)^{\frac{1}{x}}$ is differentiable at $x=0$, which is exactly showing that the limit written exists. Maybe that's obvious though and I don't see it. But it is certainly not just a composition of differentiable functions – freeRmodule Sep 11 '16 at 1:42
• You only need to calculate $f'(0)$,not $f''(0)$ – 王李远 Sep 11 '16 at 1:43
• So you're suggesting OP to find $f'(0)$ with this? $$f'(x) = \dfrac{(x+1)^{1/x} (x-(x+1) \ln(x+1))}{(x+1)x^2}$$ – David Peterson Sep 11 '16 at 1:49
• Try setting $y = (1+x)^{1/x}$and taking the natural log, then differentiating with respect to $x$. – David Bowman Sep 11 '16 at 1:49