$f(x+1)-f(x)=f'(x)$: prove $f(x)$ linear function

Suppose $f$ is differentiable, $$f(x+1)-f(x)=f'(x) \tag{1}$$ for all $x$, and $\lim_{x \to +\infty} f'(x) = A$.

I claim that $f'$ is constant, and therefore that $f$ has the form $ax+b$.

Suppose, for purposes of contradiction, that $f'$ is not constant. Then there is $x_0$ such that $f'(x_0) \ne A$. Take the case $f'(x_0) > A$. [The other case $f'(x_0)<A$ is done the same way.]

Differentiate the equation $f(x+1)-f(x)=f'(x)$ to conclude that $f''$ exists and that $f'$ is continuous. Function $f'$ achieves a maximum value $B > A$ on $[x_0,+\infty)$. The set where $f'(x)=B$ is nonempty, closed, and bounded above. Let $x_1 \in [x_0,+\infty)$ be such that $f'(x_1) = B$ and $f'(x) < B$ for all $x \in (x_1,+\infty)$.

Now note $f'(x) < B$ on $(x_1,x_1+1)$, so $f(x_1+1) - f(x_1) = \int_{x_1}^{x_1+1} f'(x)\;dx < B = f'(x_1)$. This contradicts ($1$).