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?

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.
