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

Question

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?

Answer 1

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 $-\frac{e}{2}$.

Answer 2

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*}

Answer 3

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 $e/2$.

Answer 4

$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 $\lim_{x \to 0}\dfrac{e - (1+x)^{1/x}}{x} = \dfrac{e}{2}.$

Answer 5

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 $$ e/2-xK<\frac {e-(1+x)^{\frac {1}{x}}}{x}<e/2.$$

Answer 6

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 : $f(0)=e$