TITLE: Limit of implicit function
QUESTION [3 upvotes]: For $v>0$, let $f(v)$ be the smallest positive solution $x$ of
$$\sqrt{\left(\frac{v}{x}\right)^2-1}=\tan x.$$
It can be confirmed graphically that $f(v)$ exists for all $v>0$. How can I show that
$$ \lim_{v \to \infty} f(v) = \frac{\pi}{2}?$$

REPLY [1 votes]: If $x$ is the smallest positive solution to $\sqrt{\left(\frac{v}{x}\right)^2-1}=\tan x$, then $0\le x \le \frac{\pi}{2}$ and $\left(\frac{v}{x}\right)^2-1=\tan^2 x$, from which we know that $\frac{v}{x} =\sec x$. Thus $f(v)$ = $g^{-1}(v)$, where $g(x)=x\sec(x)$. The graph of $y=g(x)$ is positive and increasing on $0\le x \le \frac{\pi}{2}$ and approaches a vertical asymptote at $x=\pi/2$, so the graph of $y=g^{-1}(v)$ is increasing with a horizontal asymptote at $y=\pi/2$.
{"set_name": "stack_exchange", "score": 3, "question_id": 981926}