If we say that $f(x)={x + 1\over x}$ and $g(x)=\sqrt[x]{x}$. It is obvious that: $$\lim_{x\to\infty} f(x)= \lim_{x\to\infty}g(x)=1$$ But, my question is: which one of these functions tends to $1$ the slowest?
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1$\begingroup$ Using the change of variable $y = \frac{1}{x},$ we want to compare $1 + \frac{1}{x} = 1 + y$ with $x^{\frac{1}{x}} = \left({\frac{1}{y}}\right)^y$ as $y \rightarrow {0}^{+}.$ So expand $\left({\frac{1}{y}}\right)^y$ in a Taylor series about $y = 0$. $\endgroup$ – Dave L. Renfro Jan 24 '20 at 16:37
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Notice that, for all $x > 0$, $$\frac{x+1}x = 1 + \frac 1 x, \qquad x^{1/x} = \exp\left(\frac{\ln x}x \right) = 1 + \frac{\ln x}{x} + \frac{(\ln x)^2}{ 2x^2} + \mathcal O(x^{-3}) $$ Ignoring all terms of order $x^{-2}$ in the Taylor series in the second equality (they are all positive, so they don't hinder our reasoning), for all $x > e$ we have $\ln x > 1$, so $$ \frac{\ln x}{x} > \frac 1 x, $$ implying $f(x) < g(x)$ for all $x > e$. As both $f$ and $g$ are monotonic on $[e,\infty)$, we may conclude that $1 < f(x) < g(x)$, that is, $f(x)$ will always be closer to $1$ than $g(x)$ for large enough $x$.
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$\begingroup$ The answer to that question seems fairly exhaustive to me. Also, take a look at the definition of Landau's small-$o$ and big-$O$ symbols to better understand how functions are classified by their growth rate $\endgroup$ – giobrach Jan 24 '20 at 16:40
