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$$\lim_{x\to\infty}\left(x^{\frac{1}{x+6}}-1\right)$$

I've been trying to solve the limit for quite some time, though I keep coming up short. I've figured that maybe it is possible to use Puiseux series, but I can't apply it for now. Can anyone break down that method or, perhaps, suggest something simpler?

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  • $\begingroup$ observe that $x^{1/x}=e^{\frac1x\ln x}$ $\endgroup$ – Masacroso May 13 '17 at 19:57
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Use the fact that $$ x^{\frac{1}{x+6}} = e^{\frac{\ln x}{x+6}}$$ and then apply l'Hopital's rule.

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For $x>0$,

$$x^{\frac {1}{x+6}}-1=e^{\ln (x)\frac {1}{x+6}}-1$$

$$=e^{\frac{x}{x+6}\frac {\ln (x)}{x}}-1. $$

using fact that $$\lim_{+\infty}\frac {\ln (x)}{x}=0$$ and $$\lim_{+\infty}\frac {x}{x+6}=1,$$

we find your limit $e^0-1=0$

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\begin{align*} \lim_{x\to\infty}\left(x^{\frac{1}{x+6}}-1\right) &= \lim_{x\to\infty}x^{\frac{1}{x+6}} - 1\\ &= \exp\left(\lim_{x\to\infty}\frac{\ln x}{x+6}\right)-1\\ &= \exp\left(\lim_{x\to\infty}\frac{1}{x}\right)-1\quad\text{(L'Hospital's Rule)}\\ &=\exp(0) - 1\\ &=1 - 1\\ &=0. \end{align*}

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