# A question regarding the asymptotic behavior of two functions

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?

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 $$x$$.
• 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 – giobrach Jan 24 '20 at 16:40