Prove that a function is uniformly continuous - without L'Hôpital's rule 
Let $f: [0,∞)→[0,∞)$ be a differentiable function, such that $ \lim\limits_{x\to\infty}\big(f(x) + f'(x)\big) = 5$. Prove that $f$ is uniformly continuous.

Now, both my tutor and my friends have proved this with a "trick" - multiplying and dividing by $ e^x $, and then using L'Hôpital for infinity over infinity.
Is there a different way of proving this?
 A: Consider $f(x)-5$ instead of $f$ so that the given limit is $0$ instead of $5$, just to simplify things.
By the limit, there exists $x_0$ such that $|f(x)+f^\prime(x)|<1$ for all $x\ge x_0$.
Suppose $f$ is not bounded from above. Then there is a sequence
$x_0<x_1<x_2<\cdots \to \infty$ with $1<f(x_1)<f(x_2)<\cdots \to \infty$.
By the MVT, there must be intermediate $y_n$ with $x_n<y_n<x_{n+1}$ and $f^\prime(y_n)>0$ and hence $f(y_n)<1$.
On $[y_{n-1},y_n]$, $f$ attains its maximum not at the boundary because alread $f(x_n)$ is bigger. So the max is attained at some interior point $z_n$.
Then $f^\prime(z_n)=0$, hence $f(x_n)\le f(z_n)\le1$, contradiction.
We conclude that $ f$ is bounded from above. By the same argument, $f$ is bounded from below. Then $f^\prime$ is also bounded on $[x_0,\infty)$, hence $f$ is Lipschitz there and consequently uniformly continuous on $[0,\infty)$.
A: We will prove if $f(x)+f'(x)\to 0$ then $f(x)\to 0.$ We can then apply this to $f(x)-5$ in the original problem.
This is a little complicated, and Hagen's answer to show $f$ is bounded is a quite a bit easier, and is enough for uniform continuity.
We will assume that it is not true that $f(x)\to 0.$
So, there is an $\epsilon>0$ such that the set $X=\{x\mid |f(x)|\geq \epsilon\}$ is unbounded.
Let $U=\{x\mid |f(x)|>\frac\epsilon2\}=f^{-1}((-\infty,\epsilon/2)\cup(\epsilon/2,+\infty)).$ Then $U$ is an open, since $f$ is continuous, and unbounded, since $X\subset U.$
Now, since $f(x)+f'(x)\to 0,$ there is an $N$ such that, for $x>N,$ $f(x)+f'(x)\in(-\epsilon/4,\epsilon/4).$ For any $x\in U$ with $x>N,$ we get $|f'(x)|>\frac\epsilon 4,$ and $f'(x)$ has the opposite sign of $f(x).$
Let $U_0=U\cap(N,\infty).$ Then $U_0$ is still unbounded and open, and for every $x\in U_0,$ $|f(x)|>\epsilon/2,$ $|f'(x)|>\epsilon/4$ and $f(x),f'(x)$ have opposite signs.
Now, if $U_0$ contains some interval $(x_0,\infty).$ Then, either,

*

*For all $x>x_0,$ $f(x)>\epsilon/2,$ or

*for all $x>x_0,$ $f(x)<-\epsilon/2.$
For $x>x_0,$ by the mean value theorem, $$f(x)=f(x_0)+(x-x_0)f'(c)$$ but we know
$|f'(c)|>\frac\epsilon4$ and has the opposite sign of $f(x_0),$ which
would mean we'd eventually get to an $x>x_0$ such that $f(x)$ is the
opposite sign, and thus, by the intermediate value theorem, an
$x>x_0$ where $f(x)=0,$ reaching a contradiction.
So $U_0$ is an unbounded disjoint union of bounded open intervals $(a_i,b_i).$ Pick any $a_i>N.$ The sign of $f(x)$ is constant in this interval. But then $f(a_i)=f(b_i)=\pm\frac\epsilon2,$ or else one of $a_i, b_i$ would be in $U_0.$ By the mean value theorem, there is a $c\in(a_i,b_i)\subset U_0$ such that $f'(c)=0.$  But $f'(c)\neq 0$ in $U_0.$
Contradiction.
So $f(x)\to 0$ and $f'(x)\to0.$

A direct proof.
Given $\epsilon>0$ there is some $N$ such that, for all $x>N,$ $|f(x)+f'(x)|<\frac\epsilon4.$
Is it possible for $f(x)> \epsilon/2$ for all $x>N?$ Then $f'(x)<-\epsilon/4$ and for all $x>N$ $$f(x)=f(N)+(x-N)f'(c)\leq f(N)-(x-N)\frac\epsilon4\to-\infty.$$
Similarly, it is not possible for $f(x)<-\epsilon/2$ for all $x>N.$
So there must be and $x_0>N$ such that $|f(x_0)|\leq \epsilon/2.$
Now, if $x>x_0,$ and $|f(x)|>\epsilon/2.$ Find the largest open interval $(a,b)$ containing $x$ such that $|f|((a,b))\in(\epsilon/2,\infty).$ Then since $x_0$ can't be in the interval, $a\geq x_0$ and $b<+\infty$ because of what we just proved.
We easily see that $f(a)=f(b)=\pm \frac\epsilon2,$ or else we could pick a larger interval including one of the $a,b$ Thus there must be a $c\in(a,b)$ such that $f'(c)=0.$ But then $|f(c)|=|f(c)+f'(c)|<\epsilon/4$ since $c>a\geq x_0>N,$ contradicting our assumption about the elements of the interval.
So for all $x>x_0,$ $$|f(x)|\leq\epsilon/2<\epsilon.$$
I suppose this is still an indirect proof. I just pulled the indirect assumption inside the proof, rather than assuming the limit doesn't exist.
