If $\lim_{x\to\infty}(f(x)+f'(x))=L$ show that $\lim_{x\to\infty} f(x) = L$ and $\lim_{x\to\infty} f'(x) = 0$ 
Let $f:(0,\infty) \to R$ be differentiable. Suppose that $\lim_{x\to\infty}(f(x)+f'(x))=L$. Show that $\lim_{x\to\infty} f(x) = L$ and $\lim_{x\to\infty} f'(x) = 0$. (Hint: Write $f(x) = e^xf(x)/e^x$ and use l’Hopital’s Rule.) 

My working for $\lim_{x\to\infty}f'(x)=0$:
For $\lim_{x\to\infty}f'(x) = 0$, I let $f(x) = e^xf(x)/e^x$ and applied quotient rule which then cancels off $e^{2x}$ and I'm left with $\lim_{x\to\infty}(f(x)+f'(x)-f(x))$. Can I then equate this with $\lim_{x\to\infty} \left(f(x)+f'(x)\right) - \lim_{x\to\infty}f(x)$ which then gives $L-L=0$? Is this step correct?
 A: You have function $f$, and it's given that $\lim_{x \to \infty} (f(x) + f'(x)) = L.$
We'll assume that $\lim_{x \to \infty} f'(x) = 0.$
If $\lim_{x \to \infty} f'(x) = 0.$, then we can rewrite:
$$\lim_{x \to \infty} (f(x) + f'(x)) = L$$ as  $$\lim_{x \to \infty} (f(x) + 0) = \lim_{x \to \infty} f(x) = L.$$
If we'll prove that $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} (f(x) + f'(x))$, then we've proved that $f'(x) = 0$.
Using the hint you gave, we'll turn $f(x)$ into $\frac{e^x f(x)}{e^x}$ and using LHopital's rule that says:
$$\lim_{x \to \infty} \frac{g(x)}{h(x)} = \lim_{x \to \infty} \frac{g'(x)}{h'(x)}$$. 
In our case, $$g(x) = {e^xf(x)} \text{ and } h(x) = e^x. $$
Let's start:
$$1. \text{Turn to suggested form:} \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{e^x f(x)}{e^x}.$$
$$2. \text{Apply LHopital's rule: differentiate both numerator and denumerator}:$$
$$ \lim \frac{(e^xf(x))'}{(e^x)'} = \lim = \frac{e^x f(x) + e^x f'x}{e^x} = \lim (f(x) + f'(x)).$$
We've got that $f(x) = f(x) + f'(x)$ that meets out assumation.
Therefore, $f'(x) = 0$.
Proven.
