Show that $\lim\limits_{x\to\infty}\frac{h(x)}{x}=c$ 
Let $a\in \mathbb{R}\cup \{-\infty\}$ and $h \colon (a,\infty)\to \mathbb{R}$ continuously differentiable. Furthermore $$\lim\limits_{x\to\infty}h'(x)=:c\in  [-\infty,\infty]$$ exists. Show that $\lim\limits_{x\to\infty}\frac{h(x)}{x}=c$.

Hint: Observe $x_n:=a+n$ for $n\in \mathbb{N}$ and $x>x_n$ and use the mean value theorem to find a relation between $h$ and $h'$. As $n\to \infty$, check $c=\infty$, $c=-\infty$ and $-\infty <c <\infty$.
Questions: If $a\in \mathbb{R}\cup \{-\infty\}$, is $x_n$ even well-defined? Couldn't I say $a=-\infty$? And what is $x_n$ for? As $n\to 0$ the sequence converges to $a$. So there is another sequence $\xi_n$ that is between $x_n$ and $a$. So we can use the mean value theorem:$$h'(\xi_n)=\frac{h(x_n)-h(a)}{x_n-a}$$ But why is that of any use?
 A: First, we should note that the hint is misleading, because it is not enough only to consider sequences of the type $x_n = b +n$ for some fixed $b \in (a,\infty)$. For example $f(x_n) = \cos(2\pi b)$, where $$f(x) = \cos(2\pi x).$$ However, this doesn't imply that $\lim_{x \rightarrow \infty} f(x)$ exists. In fact, taking $x_n = n/2$ shows that $f(x_n) = (-1)^n$.

Either you have to consider general sequences $(x_n)_{n  \in \mathbb{N}}$ with $x_n \rightarrow \infty$ or the usual $\varepsilon$-definition as used in the next argument.

We need to consider two cases: If $c \in \{\pm \infty\}$ we may w.l.o.g. that $c= \infty$. For $K>0$ let $y >0$ such that for all $x \ge y$ we have $h'(x) \ge K$. Then we find for all $x \ge \max\{2y,h(y)\}$ that
$$\tag{1}\frac{h(x)}{x} \ge \frac{h(x)-h(y)}{x} - 1  = \frac{1}{x} \int_y^x h'(t) \, \mathrm{d} t -1 \ge \frac{(K-1)}{2}.$$ 
Thus, we find that 
$$\lim_{x \rightarrow \infty} \frac{h(x)}{x}=\infty$$
because we have shown that the sequence is unbounded. As an alternative we may use the mean-value theorem to find an $\xi \in(y,x)$ such that
$$\frac{h(x)-h(y)}{x} = h'(\xi) \frac{x-y}{x} \ge \frac{K}{2}.$$
If $c$ is finite, we can find for any $\varepsilon >0$ an $y>0$ such that for any $x \ge y$ we have $|h'(x) - c| < \varepsilon$. Now you should proceed similiar as in the first case: Take $x \ge x_0 := \max  \{y, h(y)/\varepsilon, |yc|\varepsilon \}$ to find that
$$\tag{2}\left| \frac{h(x)}{x}- c \right| \leq \varepsilon  + |c|\frac{|y|}{|x|} + \left|\frac{h(x)-h(y)}{x}- c \frac{x-y}{x} \right|.$$
The choice of $x_0$ shows that $|cy|/|x| \leq \varepsilon$ and
$$\frac{h(x)-h(y)}{x} -c \frac{x-y}{x} = (h'(\xi) -c) \frac{x-y}{x}.$$
The absolute value of this guy is at most $\varepsilon$. Thus, taking all together, we find that (2) is bounded by $3 \varepsilon$ for all $x \ge x_0$. This means that
$$\lim_{x \rightarrow \infty} \frac{h(x)}{x}=c.$$

If $f$ is continuously differentiable with bounded derivate, then the existence of the limes $\lim_{x \rightarrow \infty} f(x)$ is equivalent to the convergence $f(n+b) \rightarrow L$ for $n \rightarrow \infty$ and arbitrary $b  \in (a,\infty)$, where $L \in \mathbb{R}$ is independent of $b$. 

In fact, let $(x_n)_{n \in \mathbb{N}}$ be a sequence such that $x_n \rightarrow \infty$. Then we can decompose $x_n = a_n + m_n$ with $a_n \in [0,1]$ and $m_n \in \mathbb{N}$. Now any subsequence of $(a_{n})_n$ has an convergent subsequence (by Bolzano–Weierstrass theorem), say $(a_{n_k})_k$ with limes $a \in [0,1]$. Now we have
$$|f(x_{n_k}) - f(a+m_{n_k})| \le \max |f'| |a - a_{n_k}|.$$
Together with the fact that  $ f(a+m_{n_k}) \rightarrow L$, we see that $f(x_{n_k}) \rightarrow L$. Since $L$ is independent of the subsequence, this already implies the convergence. (That is a well-known criterion for convergence in metric spaces.)
A: Given $\epsilon>0$ there is some $R$ such that $c-\epsilon<h'(x)<c+\epsilon$ whenever $x>R$. Thus, for any $a>R$ we have
$$(c-\epsilon)(x-a) < \int_a^x h'(t) \, dt = h(x) - h(a) < (c+\epsilon)(x-a),$$
Dividing with $x$ gives the inequality
$$(c-\epsilon)\frac{x-a}{x} < \frac{h(x)}{x} - \frac{h(a)}{x} < (c+\epsilon)\frac{x-a}{x}.$$
Taking limits as $x\to\infty$ results in
$$c-\epsilon \leq \lim_{x\to\infty}\frac{h(x)}{x} \leq c+\epsilon.$$
Since this is valid for all $\epsilon$ we have
$$c \leq \lim_{x\to\infty}\frac{h(x)}{x} \leq c,$$
i.e.
$$\lim_{x\to\infty}\frac{h(x)}{x} = c.$$
