To solve for A (assuming the limit exists):
$$\lim_{x\to0} \frac{60+h(x)+x+2}{-30+h(x)}=A.$$
How to evaluate this limit? Are we allowed to use the fact that, if so, then why? (definition explanation)
$$\lim_{x\to0} \:h(x)=0$$
And plug in $h(x)$ as $0$ so we get
$$\lim_{x\to0} \frac{60+x+2}{-30}=A.$$
And proceed to solve this with plugging $x=0$ since the function is continuous.
As we first plug in $h(x)=0$, we don't know if the whole function is continuous or any other information, so are we allowed to do it? Moreover, are we allowed (by the definition) to "solve" the limit by "plugging" the values more than once, namely: First $h(x)$ and then $x$.
Limit of a sum is the sum of the limits and limit of a ratio is the ratio of the limits provided the latter is non-zero. So, assuming that $\lim_{x\to 0} h(x)=0$, you can justify what you have done. The answer is $-\frac {62} {30}$.