Demostrando $\lim_{x\rightarrow 1}x^{\frac{1}{n-x-x^2-x^3-\ldots -x^n}}=\frac{1}{\sqrt[k]{e}}$

Question

Demostrando

$$\lim_{x\rightarrow 1}x^{\left(\frac{1}{n-x-x^2-x^3\cdots -x^n}\right)}=\frac{1}{\sqrt[k]{e}}$$

al $$k=1+2+3+\ldots+n$$

Answer 1

Sugerencia: $$x^{\left(\frac{1}{n-x-x^2-x^3\cdots -x^n}\right)}=\exp\left(\ln x^{\left(\frac{1}{n-x-x^2-x^3\cdots -x^n}\right)}\right)=\exp\left(\frac{\ln x}{n-x-x^2-\ldots-x^n}\right)$$ and now due to continuity of $\exp$ it suffices to show with L' Hopitals rule that $$\frac{\ln x}{n-x-x^2-\ldots-x^n} \to -\frac{1}{1+2+3+\ldots+n}=-\frac{1}{k}$$ as $x\a 1$.

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De hecho, $$\begin{align*}\lim_{x \to 1} \frac{\ln x}{n-x-x^2-\ldots-x^n}&\overset{\frac{0}{0}}=\lim_{x\to 1}\frac{(\ln x)'}{(n-x-x^2-\ldots-x^n)'}\\&=\lim_{x \to 1}\frac{\frac1x}{-1-2x-3x^2-\ldots-nx^{n-1}}\\[0.2cm]&=\frac{1}{-1-2-3-\ldots-n}=-\frac{1}{1+2+3+\ldots+n}=-\frac{1}{k}\end{align*}$$

Answer 2

Deje $S$ ser el denominador del exponente para que $$S = n - x - x^{2} - \cdots - x^{n}$$ We have to calculate the limit $$\lim_{x \to 1}x^{1/S}$$ Let this desired limit be $L$ so that $$\begin{aligned}\log L &= \log\left(\lim_{x \to 1}x^{1/S}\right)\\ &= \lim_{x \to 1}\log x^{1/S}\text{ (by continuity of log)}\\ &= \lim_{x \to 1}\frac{\log x}{S}\\ &= \lim_{x \to 1}\frac{\log x}{x - 1}\cdot\frac{x - 1}{S}\\ &= \lim_{x \to 1}1\cdot\frac{x - 1}{S}\\ &= \lim_{x \to 1}\frac{x - 1}{n - x - x^{2} - \cdots - x^{n}}\\ &= \lim_{x \to 1}\frac{x - 1}{-(x + x^{2} + \cdots + x^{n} - n)}\\ &= \lim_{x \to 1}\frac{x - 1}{-\{(x - 1) + (x^{2} - 1) + \cdots + (x^{n} - 1)\}}\\ &= \lim_{x \to 1}\dfrac{1}{-\left\{1 + \dfrac{x^{2} - 1}{x - 1} + \cdots + \dfrac{x^{n} - 1}{x - 1}\right\}}\\ &= \dfrac{1}{-\left\{1 + 2 + \cdots + n\right\}}\\ &= -\frac{1}{k}\end{alineados}$$ We have used the standard limit $$\lim_{x \to a}\frac{x^{m} - a^{m}}{x - a} = ma^{m - 1}$$ with $a = 1$ and $m = 2, 3, \ldots, n$. It now follows that the desired limit is $$L = e^{-1/k} = \frac{1}{\sqrt[k]{e}}$$