For the uniform [0,1] R.V, the mgf is $M(s)=\frac{e^s-1}{s}$ and the derivative is $M'(s)=\frac{se^s-e^s+1}{s^2}$ to calculate the mean we have to take the limit $\lim_{s\to0}M'(s)$. Why is it that for this distribution we have to take the limit and cannot evaluate at 0. Whereas for other distributions we can evaluate at 0 directly. Is this an issue that arises because of swapping the derivative with the integral because $$\frac{d}{ds}\int e^{sx}f(x)dx\big|_{s=0}= \int\frac{d}{ds} e^{sx}f(x)dx\big|_{s=0}= \int xe^{sx}f(x)dx\big|_{s=0}=\int xf(x)dx $$ seems to imply that we should be able to evaulate at 0 for all distributions including the uniform. So the trouble of not being able to evaluate at 0 should not arise.
Edit: If we assume that $s\ne0$, why then are we allowed to perform the differentiation and evaluate at 0.
You can just do that. The problem is that the derivative is not what you intuitively think it is (i.e. mere substitution). Indeed, for example, $M(0)=\mathbb{E}(e^{0\cdot X})=e^0=1$ and so the function $M$ has to be defined by parts: $M(s)=\frac{e^s-1}{s}$ if $s\neq 0$ and $M(0)=1$. Similarly,
$$M'(0)=\lim_{h\to 0}\frac{M(h)-M(0)}{h}=\lim_{h\to 0}\frac{e^h-1-h}{h^2}=\frac{1}{2}.$$
Thus you also have to write $M'$ by parts, but it is still well behaved.