Inferring Limit from Inequality 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?
 A: 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 
$$
\limsup_{x\to \infty}\frac{f(x)}x=\liminf_{x\to \infty}\frac{f(x)}x=1\\
\implies \lim_{x\to \infty}\frac{f(x)}x=1
$$
$\Box$
A: This is a standard technique used in analysis and it is just one step away from the definition of limit. Let $\epsilon>0$. Then there exist a $\lambda>1$ such that $$1-\epsilon<\frac{1}{\lambda}<\lambda<1+\epsilon $$ Note that inequality in middle $1 /\lambda <\lambda $ is obvious if $\lambda >1$. Also the first inequality holds if $\epsilon\geq 1$ and we can take $\lambda<2$ so that the last inequality also holds.
When $\epsilon<1$ then we just need to choose $\lambda <\min(1+\epsilon,1/(1-\epsilon))$ and with this choice the desired inequality holds.
Now you can proceed like this. Let $\epsilon>0$ be arbitrary. Then from the argument given above we know that there is a $\lambda >1$ such that we have $1-\epsilon<1/\lambda <\lambda <1+\epsilon$. It is known that $1/\lambda <f(x) /x<\lambda $ for all sufficiently large $x$. Combining these inequalities we get $$1-\epsilon<\frac{1}{\lambda} <\frac{f(x)} {x} <\lambda <1+\epsilon $$ for all sufficiently large $x$. By definition of limit we now have $\lim_{x\to\infty} f(x) / x=1$.

The second question uses a fairly standard technique based on $\limsup, \liminf $. The idea is that if $f(x) \leq g(x) $ then this inequality (only the weak version $\leq$) is preserved by applying $\limsup, \liminf$.
You can apply $\limsup, \liminf$ to the last equation dealing with $\vartheta(x), \pi(x) $ and then get the desired result. Thus for example lets apply $\limsup$ to get $$\limsup_{x\to\infty} \frac{\vartheta(x)} {x} \leq \limsup _{x\to \infty} \frac{\pi(x) \log x} {x} \leq \frac{1}{1-\epsilon}\limsup_{x\to\infty} \frac{\vartheta(x)} {x}, 0<\epsilon <1$$ From this (along with corresponding $\liminf$ inequality) we can see that if one of the ratios $\vartheta(x) /x, (\pi(x) \log x) /x$ has a limit the other one also has the same limit.

The definition of a limit involves inequalities and thus it is natural to expect that they can be evaluated using inequalities. It is the theorems for evaluating limits which give us the impression that limits can be evaluated using some sort of algebra. But deep down its all inequalities of a very specific kind. 
