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I have solved this limit:

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

using L'Hopital's rule and series expansion. Do you have other method for solving it?

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4 Answers 4

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I doubt one can have a simple solution to this problem without using L'Hopital's Rule or infinite series. Here is one try based on the definition of $\log x$ as an integral $$\log x = \int_{1}^{x}\frac{dt}{t}$$ First we simplify the given limit $$\begin{aligned}L\,&= \lim_{x \to 0}\frac{e - (1 + x)^{1/x}}{x}\\ &= \lim_{x \to 0}\dfrac{\exp(1) - \exp\left(\dfrac{\log(1 + x)}{x}\right)}{x}\\ &= \lim_{x \to 0}\frac{e - e^{t}}{x}\text{ (putting }t = \frac{\log(1 + x)}{x}\text{ for now)}\\ &= \lim_{x \to 0}\frac{e^{t}(e^{1 - t} - 1)}{x}\\ &= \lim_{x \to 0}e^{t}\cdot\frac{e^{1 - t} - 1}{1 - t}\cdot\frac{1 - t}{x}\\ &= \lim_{t \to 1}e^{t}\cdot\lim_{t \to 1}\frac{e^{1 - t} - 1}{1 - t}\cdot\lim_{x \to 0}\frac{1 - t}{x}\\ &= e\cdot 1\cdot\lim_{x \to 0}\frac{x - \log(1 + x)}{x^{2}}\\ &= eA\end{aligned}$$ where $$A = \lim_{x \to 0}\frac{x - \log(1 + x)}{x^{2}}$$ Now to evaluate limit $A$ we need to use certain inequalities related to $\log(1 + x)$. First let $x > 0$. then we know that $$\log(1 + x) = \int_{1}^{1 + x}\frac{dt}{t} = \int_{0}^{x}\frac{dt}{1 + t}$$ Now if $0 < t < x$ we can see that $1 - t^{2} < 1 < 1 + t^{3}$ and on dividing this by $(1 + t) > 0$ we get $$1 - t < \frac{1}{1 + t} < 1 - t + t^{2}$$ Integrating this in interval $[0, x]$ we get $$x - \frac{x^{2}}{2} < \log(1 + x) < x - \frac{x^{2}}{2} + \frac{x^{3}}{3}$$ and therefore $$\frac{1}{2} - \frac{x}{3} < \frac{x - \log(1 + x)}{x^{2}} < \frac{1}{2}$$ Letting $x \to 0^{+}$ and using squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{x - \log(1 + x)}{x^{2}} = \frac{1}{2}$$ We can similarly show (using another set of inequalities) that $$\lim_{x \to 0^{-}}\frac{x - \log(1 + x)}{x^{2}} = \frac{1}{2}$$ It now follows that $A = 1/2$ and the limit $L = e/2$.

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We can find the limit: $$\lim_{n\to +\infty} n\left(e-\left(1+\frac{1}{n}\right)^n\right)$$ by exploiting the convexity of the exponential function, for which: $$x<y\quad\Longrightarrow\quad e^x<\frac{e^y-e^x}{y-x}<e^y.$$ If we take $y=1$ and $x=n\log\left(1+\frac{1}{n}\right)$ (we know that $y>x$ by the Bernoulli inequality) we have: $$\left(1+\frac{1}{n}\right)^n <\frac{e-\left(1+1/n\right)^n}{1-n\log(1+1/n)}<e$$ hence we just need to find the limit: $$\lim_{n\to +\infty}\left(n-n^2\log(1+1/n)\right)=\frac{1}{2}$$ that follows from: $$\log(1+1/n) = \frac{1}{n}-\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right)$$ to prove that the original limit equals $\frac{e}{2}$.

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Let $f(x)=(x+1)^{1/x}$. We want to compute $\lim_{x\to 0}\frac{e-f(x)}{x}$. First, $$\begin{align*} \lim_{x\to 0}\ln f(x) &= \lim_{x\to 0}\frac{\ln (x+1)}{x}\\ &= \lim_{x\to 0}\frac{\ln (x+1)-\ln 1}{x}\\ &= [\ln(x)]'(1)\\ &= 1. \end{align*}$$ Hence $\lim_{x\to 0}f(x)=e$. Therefore $$\begin{align*} \lim_{x\to 0}\frac{f(x)-e}{x}&= \lim_{y\to 0}f'(y)\\ &= \lim_{y\to 0} f(y)[\ln f]'(y)\\ &= e\lim_{y\to 0} \left(\frac{\ln(x+1)}{x}\right)'(y)\\ &= e\lim_{y\to 0} \frac{\frac{y}{y+1}-\ln(y+1)}{y^2}\\ \mbox{L'Hospital}&= e\lim_{y\to 0} \frac{\frac{1}{(y+1)^2}-\frac{1}{y+1}}{2y}\\ &= e\lim_{y\to 0} \frac{-y}{2y(y+1)^2}\\ &= \frac{-e}{2}. \end{align*}$$ Hence we've shown that $\lim_{x\to 0}\frac{e-f(x)}{x}=\frac{e}{2}$ as desired.

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But he asked for something other than L'Hopital. –  Conifold Jul 26 at 23:48

An alternative approach: Recall by Taylor Series that $\ln(x+1)=x-\frac{x^2}{2}+o(x^3)$. Thus $$\begin{align*} (x+1)^{1/x}&= \exp\left[\frac{\ln(x+1)}{x}\right]\\ &= \exp\left[1-\frac{x}{2}+o(x^2)\right]\\ &= e\exp \left[-\frac{x}{2}+o(x^2)\right]. \end{align*}$$ Thus we compute $$\begin{align*} \lim_{x\to 0}\frac{e-(x+1)^{1/x}}{x}&= e\lim_{x\to 0}\frac{1-\exp \left[-\frac{x}{2}+o(x^2)\right]}{x}\\ &= -e\left(\exp \left[-\frac{x}{2}+o(x^2)\right]\right)'_{x=0}\\ &= -e\left(-\frac{x}{2}+o(x^2)\right)'_{x=0}\left(\exp \left[-\frac{x}{2}+o(x^2)\right]\right)_{x=0}\\ &=\frac{e}{2}. \end{align*}$$

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