Limit of implicit function

Question

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 $$ \lim_{v \to \infty} f(v) = \frac{\pi}{2}?$$

Answer 1

Let $x(v)$ be the solution for each $v$. You can show such a solution exists for all positive $v$ by a number of methods, one being by looking at the function $g_v(x)=\tan(x)-\sqrt{(v/x)^2-1}$ and showing that for $v>\pi$: $g(1)>0$, $g(\pi)<0$, so by continuity a solution always exists. This also implies $0<x(v)<\pi$, meaning that it's bounded.

Now we just need to show that $x(v)$ is increasing. This is easy because looking at $g_v(x)$, for fixed $x$, increasing $v$ makes $g_v(x)$ negative, and since $\tan(x)$ blows up at half integer multiples of $\pi$, we get that $x(v)$ is increasing. You need a bit more rigor here, work it through. So $x(v)$ has a limit as $v\rightarrow\infty$.

Using the fact that $x(v)=\arctan(\sqrt{(v/x(v))^2-1})$ and that $(v/x(v))$ is going to infinity as $x\rightarrow\infty$, we get $\lim_{v\rightarrow\infty} x(v) = \lim_{z\rightarrow\infty} \arctan(z)=\pi/2$.

Answer 2

Since $x>0$, then $\tan x=\sqrt{\left(\frac{v}{x}\right)^2-1} \to +\infty$ as $v\to +\infty$. So $x=\arctan x \to \frac{\pi}{2}$ as $v\to +\infty$. And $x$ is asked to be smallest positive solution, hence $x=\frac\pi 2$.

Answer 3

Let

$$ g_v(x) = \sqrt{\left(\frac{v}{x}\right)^2-1}-\tan x. $$

Then for fixed $v$

$$ \lim_{x \to \pi/2^-} g_v(x) = -\infty, $$

and for fixed $x$

$$ \lim_{v \to \infty} g_v(x) = \infty. $$

Now use the intermediate value theorem to show that, for any fixed $\epsilon > 0$, $g_v(x)$ has a zero $x = f(v)$ in the interval $(\pi/2-\epsilon,\pi/2)$ for $v$ large enough.

Answer 4

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 $y=\pi/2$.