$x^2-\log x = u $ asymptotic behaviour

Question

Find the asymptotic behaviour as $u \to \infty$ of the solutions of $x^2-\log x = u$. Is there a standard method to solve this kind of problems? May the fact that we obviously know the derivative of $f^{-1}(x)$ help? A similar problem was: "given $u=x+\tanh x$ show that, if $u \to \infty$ then $x=u-1+e^{2-2u}+O(e^{-4u})$

Answer 1

I would do this by Newton's method. Obviously, for $x^2 - \log x = u,$ the first guess would be $x = \sqrt{u},$ (since $\log x \ll x,$ for $x$ large). The first Newton step ought to give you an excellent approximation. As for your "similar problem", what you wrote makes no sense (since there is a mix of $x$s and $u$s), so you might want to edit. That said, the solution should be very similar, since for $x\gg 1,$ we know that $\tanh x$ is very close to $1,$ so obviously $x = u-1$ is the first guess, and you can get the second by Newton.