There are two cases where I failed to understand how limit is obtained from an inequality due to my incomplete knowledge of real analysis (limit, integration), i know basic definitions but I can't put it together under the context. I am giving the cases below-
Case A: $\lim_{x \to \infty} f(x)/x=1$ can be derived from –
$$1/\lambda<f(x)/x<\lambda \cdots (1)$$
the statement is taken from below proof-
Lemma 16.8. Let $f:\Bbb R_{\geq1} \to \Bbb R$ be a nondecreasing function. If the integral $\int_1^\infty \frac{f(t)-t}{t^2} dt$ converges then $f(x)\sim x$.
Proof. Let $F(x):=\int_1^{x}\frac{f(t)-t}{t^2}dt$. The hypothesis is that $\lim_{x\to\infty}F(x)$ exists. This implies that for all $\lambda>1$ and all $\epsilon>0$ we have $|F(\lambda x)-F(x)|<\epsilon$ for all sufficiently large $x$.
Fix $\lambda>1$ and suppose there is an unbounded sequence $(x_n)$ such that $f(x_n)\geq \lambda x_n$ for all $n\geq 1$. For each $x_n$ we have
$$F(\lambda x_n)-F(x_n)=\int_{x_n}^{\lambda x_n} \frac{f(t)-t}{t^2} dt\geq\int_{x_n}^{\lambda x_n}\frac{\lambda x_n-t}{t^2}dt=\int_1^{\lambda}\frac{\lambda-t}{t^2}dt=c$$ for some $c>0$, where we used the fact that $f$ is non-decreasing to get the middle inequality. Taking $\epsilon<c$, we have $|F(\lambda x_n)-F(x_n)|=c>\epsilon$ for arbitrarily large $x_n$, a contradiction. Thus $f(x)<\lambda x$ for all sufficiently large $x$. A similar argument shows that $f(x)>\frac1{\lambda}x$ for all sufficiently large $x$. These inequalities hold for all $\lambda>1$, so $\lim_{x\to\infty}f(x)/x=1$. Equivalently, $f(x)\sim x. \quad\square$
Question: How $\lim_{x \to \infty} f(x)/x=1$ can be derived from $1/\lambda<f(x)/x<\lambda$?
I have taken it granted that $1/\lambda<f(x)/x<\lambda$. This I posted it before but could not understand the answer that was given by a forum member.
Case B: Also. I have been told in another post that –
since $\frac{\log x}{x^\epsilon}\to 0$ for any $\epsilon >0$, we can
make $\frac{\theta(x)}x$ arbitraly close to $\frac{\pi(x)\log x}{x}$.
The statement is given for below inequality –
$$\frac{1}{1-\epsilon} \frac{\theta(x)}{x}\leq \pi(x)\log(x)/x \leq \frac{\theta(x)}{x} \cdots (2)$$
The source of the inequality is given below –
Theorem 16.6 (Chebyshev). $\pi(x)\sim \frac{x}{\log x}$ if and only if $\vartheta(x)\sim x$.
Proof. We clearly have $0\leq\vartheta (x)\leq\pi(x)\log x$, thus
$$\frac{\vartheta(x)}{x}\leq\frac{\pi(x)\log x}{x}.$$
For every $\epsilon>0$ we have
\begin{align}
\vartheta(x)\geq\sum_{x^{1-\epsilon}<p\leq x}\log p &\geq (1-\epsilon)(\log x)\big(\pi(x)-\pi(x^{1-\epsilon})\big)\\
&\geq (1-\epsilon)(\log x)(\pi(x)-x^{1-\epsilon})\\
\end{align}
and therefore
$$\pi(x)\leq\bigg(\frac1{1-\epsilon}\bigg)\frac{\vartheta(x)}{\log x}+x^{1-\epsilon}.$$
Thus for all $\epsilon>0$ we have
$$\frac{\vartheta(x)}{x}\leq\frac{\pi(x)\log x}{x}\leq \bigg(\frac1{1-\epsilon}\bigg)\frac{\vartheta(x)}{x}+\frac{\log x}{x^\epsilon}.$$
The second term on the RHS tends to $0$ as $x\to\infty$, and the lemma follows: by choosing $\epsilon$ sufficiently small we can make the ratios of $\vartheta(x)$ to $x$ and $\pi(x)$ to $x/\log x$ arbitrarily close together as $x\to\infty$, so if one of them tends to $1$, then so must the other. $\quad\square$
Question: If we choose $x=e^y$, then $\frac{1}{1-\epsilon} \frac{\theta(x)}{x}$ becomes negative, then the gap between $\frac{1}{1-\epsilon} \frac{\theta(x)}{x}$ and $\frac{\theta(x)}{x}$ is large, then how $\pi(x)\log(x)/x$ becomes close to $\frac{\theta(x)}{x}$?
General Question: In both cases. the limit was established from inequality, but couldn't comprehend how it is done, could anyone please elaborate it how is done?
Best Answer
I just will answer your first question. Please, write just one question per post.
From the first quote of your post we knows that for each chosen $\lambda>1$ there is some $y_\lambda>1$ such that $f(x)/x<\lambda$ for all $x>y_\lambda$ and some $z_\lambda >1$ such that $f(x)/x>1/\lambda$ for all $x>z_\lambda$, thus setting $w_\lambda :=\max\{y_\lambda ,z_\lambda \}$ we find that $$ \forall \lambda >1,\exists w_\lambda >1,\forall x>w_\lambda : \frac1\lambda<\frac{f(x)}x<\lambda\tag1 $$ Then from $\rm (1)$ taking limits we can see that $$ \forall \lambda >1:\limsup_{x\to\infty}\frac{f(x)}{x}\leqslant \lambda \\ \qquad \implies\limsup_{x\to \infty }\frac{f(x)}x\leqslant \inf\{\lambda:\lambda >1\} =1\tag2 $$ and similarly $$ \forall \lambda >1:\liminf_{x\to\infty}\frac{f(x)}{x}\geqslant \frac1\lambda \\ \qquad \implies\liminf_{x\to \infty }\frac{f(x)}x\geqslant \sup\{\tfrac1\lambda:\lambda >1\} =1\tag3 $$ And because $\liminf_{x\to\infty}g(x)\leqslant \limsup_{x\to\infty}g(x)$ for any real-valued function $g$ then $$ \limsup_{x\to \infty}\frac{f(x)}x=\liminf_{x\to \infty}\frac{f(x)}x=1\\ \implies \lim_{x\to \infty}\frac{f(x)}x=1 $$ $\Box$