$f'(t)\rightarrow b$ as $t\rightarrow +\infty$ $\Rightarrow f (t)/t\rightarrow b $ using Mean Value Theorem 
Suppose that $f $ is differentiable on $(0,\infty) $ and $f'(t) \rightarrow b $ as $t\rightarrow + \infty$ . Show that $f (t)/t\rightarrow b $.

I want to use mean value theorem here, but I can't figure out where. When $ f (0) \ne 0$, I can't make $f'(c)=\frac {f (t)-f (a)}{t-a} $ become $f (t)/t $. So now I am just not really sure what to do.
 A: Given an $\epsilon>0$ there is an $M>0$ such that
$$|f'(t)-b|<{\epsilon\over2}\qquad(t>M)\ .$$
Given any $t>M$ the MVT guarantees the existence of a $\xi>M$ with
$$f(t)-f(M)=f'(\xi)(t-M)=\bigl(f'(\xi)-b\bigr)(t-M) + b(t-M)\ .$$
It follows that $f(t)-bt=\bigl(f'(\xi)-b\bigr)(t-M)+f(M)-bM$, so that
$$\bigl|f(t)-bt\bigr|\leq{\epsilon\over2}(t-M)+\bigl|f(M)\bigr|+|b|M\qquad(t>M)\ .$$
After dividing by $t$ we therefore obtain
$$\left|{f(t)\over t}-b\right|\leq{\epsilon\over2} +{\bigl|f(M)\bigr|+|b|M\over t}\qquad(t>M)\ .$$
Here the right side is $<\epsilon$ as soon as $t>M'$ for a suitable $M'>M$.
A: Note: I've avoided using $\limsup$ and $\liminf$, so the proof is more complicated as a result.
Let $\epsilon > 0$ be given. You want to show that there is a number $M$ such that for all $t \geq M$ you have 
$$ b-\epsilon \leq f(t)/t \leq b + \epsilon. $$
Since $f'(t) \to b$, there is an $A$ such that whenever $t \geq A$, you have
$$b - \epsilon/2 \leq f'(t) \leq b + \epsilon/2.$$
Now let $g(t) = f(A) + (b + \epsilon/2)(t-A) - f(t)$. Then we have $g(A) = 0$. Since $g'(t) = b + \epsilon/2 - f'(t) \geq 0$ for $t \geq A$, the function $g$ must be nondecreasing on $[A,+\infty)$, hence $g(t) \geq 0$ for $t \geq A$. Thus for all $t \geq A$ we have 
$$f(t) \leq f(A) + (b + \epsilon/2)(t-A).$$
Letting $M_1 = \sup[A,(2/\epsilon)(f(A) - (b + \epsilon/2)A)]$, then for $t \geq M_1$, the right-hand side is bounded above by  $(b + \epsilon)t$, showing that $f(t)/t \leq b + \epsilon$ whenever $t \geq M_1$.
Using the function $h(t) = f(t) - f(A) - (b-\epsilon/2)(t-A)$, it can similarly be proved that there is an $M_2$ such that for all $t \geq M_2$, we have $f(t)/t \geq b-\epsilon$.
Now it suffices to take $M = \sup(M_1,M_2)$.
EDIT: Here's a simpler proof using $\limsup$. 
Let any number $\epsilon > 0$ be given. Then there is some $A$ such that $f'(t) \leq b + \epsilon$ whenever $t \geq A$. Then by the mean value theorem, we must have $[f(x) - f(A)]/(x-A) \leq b+\epsilon$ for all $x > A$. (Argue by the absurd.)  So $f(x) \leq f(A) + (b +\epsilon)(x-A)$.
Thus 
$$f(x)/x \leq \frac{f(A) + (b + \epsilon)(x-A)}{x} \to b + \epsilon$$
as $x \to +\infty$. This proves $\limsup_{x \to +\infty} f(x)/x \leq b + \epsilon$. Since $\epsilon$ is arbitrary, this shows in fact that $\limsup_{x \to +\infty} f(x)/x \leq b$.
A similar argument shows that $\liminf_{x \to +\infty} f(x)/x \geq b$, and the conclusion follows.
