Prove $ \lim_{h\rightarrow 0}\frac{1}{h}\int_a^x[f(t+h)-f(t)]\mathrm{d}t=f(x)-f(a). $

This is a simple consequence of the Fundamental Theorem of Calculus. Also you need to have $h\to 0^{+}$ instead of $h\to 0$ (why? Explain yourself).

Clearly if $F$ is an anti-derivative of $f$ on $[a, b] $ (such an $F$ exists because of continuity of $f$ and Fundamental Theorem of Calculus) then we have $$\int_{a}^{x}f(t)\,dt=F(x)-F(a)$$ and $$\int_{a} ^{x} f(t+h) \, dt=F(x+h) - F(a+h) $$ and thus the desired limit equals the limit of expression $$\frac{F(x+h) - F(x)} {h} - \frac{F(a+h) - F(a)} {h} $$ and since $F$ is the anti-derivative of $f$ the desired limit equals $$F'(x) - F'(a) =f(x) - f(a) $$