Why are $dT^2$ and $dT^3$ negligibly small? With $dT^2$ I mean the square of the change in temperature. While deriving the relationship between the coefficient of linear expansion and of volumetric expansion, terms with $dT^2$ and $dT^3$ are said to be ignored because they are very small. 
Can anyone explain why they are ignored when temperature change can be bigger?
 A: Because $dT$ is a tiny -  almost infinitely tiny - value. A tiny, tiny change in temperature. You say that "temperature changes can be bigger", and that is true, but then they will not be called $dT$ (but rather $\Delta T$). When you see the notation $dT$ you know that you have something infinitesimally small. 
When you multiply something tiny with something tiny, it becomes even smaller. Just think of squaring and cubing a value such as $0.2$:
$$0.2^2=0.04\qquad 0.2^3=0.008$$
It becomes smaller and smaller. $dT^2$ and $dT^3$ are seriously very small. The extra expansion that they cause on top of it all is so small that you maybe can't even measure it. 
And therefore people have decided to disregard them, because that makes the formula so much simpler to work with. The result with the simplified formula is a tiny bit off, but that should be almost nothing. 
A: Only in the infinitesimal limit one can disregard higher order terms. In all cases when we are consider finite differences of temperature and we are approximating changes in functionals based on the local values of its partial derivatives, we have to include higher order terms until the required precision. This is how Taylor expansions work
