What is the purpose of dividing by $\delta t$ in the definition of the hazard function? My lecturer explained that it is there so that the limit doesn't go to zero and that it acts as some sort of 'normalising constant'. Any explanation would be welcome.
The definition is: $h(t) = \lim_{\delta t \downarrow 0} \frac{P(t \leq T < t+ \delta t |T \geq t)}{\delta t}$
I presume $T$ is a random variable representing a failure time. And your equation should probably read $$h(t) = \lim_{\delta t\downarrow 0} \frac{P(t \le T < t+\delta t\mid T\ge t)}{\delta t}.$$
The hazard function represents an instanteous rate of failure. So if we're at time $t$ and it hasn't failed and we look out into the immediate future, the numerator represents the probability that the failure will happen within time $\delta t.$
The bigger $\delta t$ is, the more time there is for the failure to occur, so the probability of failure within time $\delta t$ is going to be bigger when $\delta t$ is big and $smaller$ when it's small. If $\delta t$ is vanishingly tiny, the probability of failure will be vanishingly tiny as well. In fact in the limit $\delta t\rightarrow 0$ we might expect that the probability vanishes proportional to $\delta t.$
Thus the hazard function tells you the probability of failure per time, expressed as an instantaneous rate.