If $\lim_{x\to\infty}(f(x)+f'(x))=L$ show that $\lim_{x\to\infty} f(x) = L$ and $\lim_{x\to\infty} f'(x) = 0$

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.