Prove that If $f(x)\to a$ and $f'(x) \to b$ as $x\to +\infty$, then $b = 0$ In the book of The elements of Real Analysis, by Bartle, at page 220, it is asked to show that

If $f(x)\to a$ and $f'(x) \to b$ as $x\to +\infty$, then $b = 0$

However, I'm having trouble showing the result.
By our assumption, for a given $\epsilon > 0$, $\exists M \in \mathbb{R}$ and $\delta >0 $ such that $\forall x > M$, $$|f(x) - a| < \epsilon,$$ and 
$$|\frac{f(x+h) - f(x)}{h} - b| < \epsilon$$ for $0 < |h| < \delta.$
However, (by observing $x+h > x > M$), I get 
$$-2\epsilon / h - b<|f'(x) - b| < 2\epsilon / h - b,$$ which does not let me do anything because of that $h$ in the denominator of $\epsilon$. So my question is, how can we prove this result?
 A: Correct me if wrong :
1)$ \lim_{x \rightarrow \infty} f(x) =a$:
Let $\epsilon$ be given .
There exists an $M$, real, such that for $x\ge M $
$|f(x)-a| \lt \epsilon.$
2) MVT:
Consider $h \gt 0$, $h$ fixed.
$hf'(t) = f(x+h) - f(x)$, with
$x \lt t \lt x+h$.
Let $x \gt M$.
$h|f'(t)| = |f(x+h) -f(x)| =$
$ |(f(x+h) -a) -(f(x) -a)| \le$
$|f(x+h) -a| +|f(x)-a| \lt 2\epsilon.$
Recall : $x \lt t\lt x+h:$
$ x \rightarrow \infty$ implies $t \rightarrow \infty$ :
$h \lim_{t \rightarrow \infty} |f'(t)|= h|b| \le 2\epsilon.$
This  implies;
$\lim_{t \rightarrow \infty}|f'(t)|=|b|= 0.$
Note: If $|b| \not=0$ we get a contradiction by choosing $h$, an independent parameter, sufficiently large.
A: Just to expand the diversity of the answers, I'm going to post the proof suggested by @ParamanandSingh
Let for a fixed $x\in \mathbb{R}$, consider the interval $(x, x+1)$. By MVT, $\exists c \in (x, x+1)$ s.t
$$\frac{f(x+1) - f(x)}{1}  = f'(c).$$
Now if we let $x\to +\infty$, $c\to +\infty$, and the difference $$f(x+1) - f(x) \to 0,$$ hence
$$f'(c) \to b\quad as \quad c\to +\infty$$
implies $$b = 0$$
QED.
A: If we assume
$f'(x) \to b \; \text{as} \; x \to \infty, \tag 1$
then we have, for any $\epsilon > 0$, a sufficiently large $M \in \Bbb R$ such that
$b - \epsilon < f'(x) < b + \epsilon \; \text{for} \; x \ge M; \tag 2$
suppose that $b > 0$; then we choose $\epsilon$ so small that
$b - \epsilon > 0; \tag 3$
then
$f(x) - f(M) = \displaystyle \int_M^x f'(s) \; ds \ge \int_M^x (b - \epsilon)\; ds = (b - \epsilon)(x - M), \tag 4$
whence
$f(x) \ge f(M) + (b - \epsilon)(x - M) \to \infty \; \text{as} \; x \to \infty; \tag 5$
likewise, if $b < 0$, we may choose $\epsilon$ such that
$b + \epsilon < 0; \tag 6$
then by an argument similar to the above we have
$f(x) \le f(M) + (b + \epsilon)(x - M) \to -\infty \; \text{as} \; x \to \infty; \tag 7$
in neither case $b > 0$, $b < 0$ does $f(x) \to a$ as $x \to \infty$; thus, we must have
$b = 0. \tag 8$
A: Edit:
As it is pointed out, there is a mistake in this proof, but it can be solved by observing that the difference $f(x+1)- f(x)$ can be arbitrarily small for sufficient large values of $x$.

After getting some ideas, it is clear that easiest method is to use the method of contradiction, so I'm going to post my own proof in here.
WLOG, let $f'(x) > 0$ as $x\to +\infty$, then by the lemma $19.3.a$ in Bartle's book, saying that, $\exists h> 0$ s.t for all y satisfying $x< y < x+h$, we have $f(x)< f(y)$. Then let $f(y) - f(x) = \epsilon'$ and choose $2\epsilon < \epsilon'$ so that $\exists M$ and $\forall x > M$, we have 
$$|f(x) - a| <  \epsilon.$$
Since $y > x > M,$ $|f(y)- a| < \epsilon$, hence
$$0 < \epsilon' < |f(y)- f(x)| < 2\epsilon$$, which is a contradiction.
QED
