$$\displaystyle \lim_{x \to \infty} x^{1/x}$$ It's supposed to be calculated with L´Hospital rules, but I can´t find the way of express the limit as a quotient.
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1Use $a^b = e^{b\cdot \log a}$ to get your quotient. But L'Hospital's rule is unnecessarily complicated. – Daniel Fischer Apr 03 '14 at 15:43
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Write you limit as $y=...$ and take the natural log on both sides. That $1/x$ term then "comes out" of the exponent and arrives as an $x$ in the denominator. Try it out form there. – imranfat Apr 03 '14 at 15:44
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L'Hospital? What for? $\log x\to-\infty$, $1/x\to+\infty$, where is the indeterminate form? The thought-killing machine seems to be working at full power here... – Did Apr 03 '14 at 15:44
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The text of the question has just been changed (without any mention of the fact...). As a consequence, the limits stated in my previous comment do not apply anymore. – Did Apr 03 '14 at 15:51
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EDIT since the question changed: $x^{1/x} = e^{\log(x^{1/x})} = e^{\frac{\log x}{x}}$. Then you have:$$\lim_{x \to \infty}x^{1/x} = \lim_{x \to \infty}e^{\frac{\log x}{x}} = e^{\lim_{x \to \infty}\frac{\log x}{x}} = 1,$$ where the last equality follows noticing that $\lim_{x \to \infty}\frac{\log x}{x} = 0$.