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Assuming both limits exist, apply L’Hospitals rule:
$$L= \lim_{x \rightarrow \infty}f(x) = \lim_{x \rightarrow \infty}\frac{e^xf(x)}{e^x}\\=\lim_{x \rightarrow \infty}[f(x) + f'(x)] = L + \lim_{x \rightarrow \infty} f'(x) , $$
and conclude that $ \lim_{x \rightarrow \infty} f'(x)= 0$.
Alternatively, by the MVT there is a point $\xi_x \in(x,x+1)$ such that
$$ f(x+1)-f(x)=f'(\xi_x)$$
If $f'(x) \rightarrow L’$, then for every $\epsilon > 0$ there is a $K>0$ such that $|f'(x) – L’| < \epsilon$ when $x > K$. As $\xi_x > x > K$, it follows that $|f'(\xi_x) – L’| < \epsilon.$
Hence,
$$ L’=\lim_{x \rightarrow \infty}f'(x)=\lim_{x \rightarrow \infty}f'(\xi_x)=\lim_{x \rightarrow \infty}[f(x+1)-f(x)]=0.$$
If $\lim_{x\rightarrow\infty}f'(x)=c$ were some positive number, that would imply, for some $0<k<c$ and all large enough $x$ that $f'(x)>k$. Think about what this means intuitively and why this is inconsistent with $f$ converging.