Asymptotics of roots of equations and implicit functions

$x(\varepsilon ) \rightarrow 0$ as $\varepsilon \rightarrow 0+$;

yet for any $s > 0$, $\varepsilon^{-s} x(\varepsilon ) \rightarrow \infty$ as $\varepsilon \rightarrow 0+$.

Prove that the equation $\operatorname {tg} x = a / x, a > 0$, has exactly one solution on each interval $(-\pi / 2+\pi n ; \pi / 2+\pi n), n \in \mathbb {N}$.

Let $x_{n}$ be the solution of the equation $\operatorname {tg} x = a / x, a > 0$, lying in the interval $(-\pi / 2+\pi n ; \pi / 2+\pi n), n \in \mathbb {N}$. Prove that

Prove that the equation $\operatorname {tg} x = a x, \quad a > 0$, has exactly one solution $x_{n}$ on each interval $(-\pi / 2+\pi n ; \pi / 2+\pi n), n \in \mathbb {N}$.

Let $x_{n}$ be the solution of the equation $\operatorname {tg} x = a x, a > 0$, lying in the interval $(-\pi / 2+\pi n ; \pi / 2+\pi n), n \in \mathbb {N}$. Prove that for all sufficiently large $n \in \mathbb {N}$

Find $\lim_{n \rightarrow \infty }\left(x_{n+1}-x_{n}\right)$, where the sequence $\left\{ x_{n}\right\}$ is defined in 1) and 2).

$x_{n} = \pi n+o(1)$;

$x_{n} = \pi n+(-1)^{n} /(\pi n)+o(1 / n)$;

$x_{n} = \pi n+(-1)^{n} /(\pi n)-\left(1 /\left(\pi^{3} n^{3}\right)\right)\left(1-(-1)^{n} / 6\right)+o\left(1 / n^{3}\right)$.

$x_{n} = \frac{\pi }{2}+\pi n+o(1)$;

$x_{n} = \frac{\pi }{2}+\pi n-\frac{1}{\pi / 2+\pi n}+o\left(\frac{1}{n}\right)$;

$x_{n} = \frac{\pi }{2}+\pi n-\frac{1}{\pi / 2+\pi n}-\frac{2}{3(\pi / 2+\pi n)^{3}}+o\left(\frac{1}{n^{3}}\right)$.

$y(x)=\ln x-\ln \ln x+c_{1} a_{1}(x)+c_{2} a_{2}(x)+O\left(\frac{a_{2}(x)}{\ln \ln x}\right)$;

$y(x)=\sqrt{x}\left(c_{0}+c_{1} a_{1}(x)+c_{2} a_{2}(x)+O\left(\frac{a_{2}(x)}{\ln x}\right)\right)$;

$y(x)=\frac{\ln x}{x}+\frac{1}{x}\left(c_{1} a_{1}(x)+c_{2} a_{2}(x)+O\left(\frac{a_{2}(x)}{\ln x}\right)\right)$;

$y(x)=\frac{x}{\ln x}\left(c_{0}+c_{1} a_{1}(x)+c_{2} a_{2}(x)+O\left(\frac{a_{2}(x)}{\ln \ln x}\right)\right)$, where $c_{0}, c_{1}, c_{2}$ are constants, and $a_{k}$ are functions of the form $\frac{(\ln x)^{m}}{x^{n}}$ or $\frac{(\ln \ln x)^{m}}{\ln^{n} x}$ for suitable $m, n \in \mathbb {N}$;

$y(x)=-x^{2}+1-e^{x-x^{3}}+O\left(x e^{2\left(x-x^{3}\right)}\right)$;

$y(x)=-\frac{x}{2}+\frac{\pi }{4}-\frac{1}{x}+O\left(\frac{1}{x^{2}}\right)$.

$y \in C^{\infty }((0, \infty )), 0<y<1, y^{\prime }>0$;

$y(x)=x\left|\ln x\right|+x \ln \left|\ln x\right|+o(x)$ as $x \rightarrow +0$;

$y(x)=1-\frac{\ln x}{x}+\frac{\ln \ln x}{x}+o\left(\frac{1}{x}\right)$ as $x \rightarrow +\infty$.

$x_{n}=\pi \left(n+\frac{1}{2}\right)-\frac{1}{\pi n}+O\left(\frac{1}{n^{2}}\right)$;

$-\frac{1}{\pi \left(n+\frac{1}{2}\right)}-\frac{1}{(\pi n)^{3}}<x_{n}-\pi \left(n+\frac{1}{2}\right)<-\frac{1}{\pi \left(n+\frac{1}{2}\right)}$.