If f′ tends to ℓ, then f(x+1) − f(x) tends to ℓ

1. For each $x$ there exists $c_x\in(x,x+1)$ with $f(x+1)-f(x)=f'(c_x)$. Since $c_x>x\to+\infty$, we have $f'(c_x)\to\ell$.

2. For $f=\sqrt\cdot$, $f'(x)=\frac1{2\sqrt x}\to0$, so $\sqrt{x+1}-\sqrt x\to0$. For $f=\ln$, $f'(x)=\frac1x\to0$, so $\ln(x+1)-\ln x\to0$.

3. No. Let $f(x)=\dfrac{\sin(x^2)}x$ for $x\ge1$. Then $|f(x+1)-f(x)|\le\frac1{x+1}+\frac1x\to0$, so the differences tend to $0$, but
$$f'(x)=2\cos(x^2)-\frac{\sin(x^2)}{x^2}$$
oscillates between nearly $-2$ and $2$: it has no limit.

Remark. Information on a derivative always gives information on differences (through the mean value theorem). The converse requires extra hypotheses.