Reformulating a discrete model into a continuous model? 
Past Paper Question:

Given a model of a ﬁsh population undergoing constant effort harvesting, where $N_i$ is the total population at time point $i$:
$$N_{i+1}=N_{i}+BN_{i}-DN_{i}^{2}-EN_{i},$$
where $BN_{i}$ are the number of births, $DN_{i}^{2}$ are the number of death, and $EN_{i}$ is the harvesting rate.
Reformulate the above model as a discrete model for constant yield ﬁshing, then take a small time limit to ﬁnd the equivalent continuous model.

My Attempt:
Not sure about the first part, but for the second part I think you make the substitution $N_{i+1}=N\left( t+\delta t\right)$ to get the form:
  $$\dfrac {N\left( t+\delta t\right) -N\left( t\right) } {\delta t}=BN-DN^{2}-EN,$$
  let $\delta t\rightarrow 0$ to give:
  $$\dfrac {dN } {dt}=BN-DN^{2}-EN$$

My Question:
Is this the correct approach and what is the final answer? 
 A: To understand what is asked in the first part, we'll take a closer look at the given model.
$$N_{i+1} = N_i + BN_i - DN_i^2- EN_i$$
Because of the $EN_i$ term, the population being harvested is proportional (by a constant $E$) to the total amount of population at time $i$, e.g. $N_i$. This is known as proportional harvesting since the amount of harvest is proportional to the total number of population.
On to the question, constant (rate) harvesting means the population being harvested at each iteration is independent (or fixed) of the total population. This means that your new harvesting term should be just $E$. Thus your constant harvesting model looks like this:
$$N_{i+1} = N_i + BN_i - DN_i^2- E$$
Note: the meaning (and unit) of $E$ in both cases are different. In the first case, $E$ represents a unit-less proportion constant - which is sometimes expanded as the product of catch-ability and effort. On the other hand, in the second case, your $E$ is the fixed amount of population being harvested, thus it has unit of whatever species being considered. 
For the second part, you are heading in the right direction. I will expand upon your attempt. Set $N_{i+1} = N(t+\delta t)$, this explicitly implies each time step is $\delta t$. To find the rate of change (or the differential equation) then subtracting $N_(t)$ from both sides:
$$N(t+\delta t) - N(t) = B N(t) - D N(t)^2 - E$$
Divide both sides by $\delta t$ to obtain:
$$\frac{N(t+\delta t) - N(t)}{\delta t} = \frac{B}{\delta t} N(t) - \frac{D}{\delta t} N(t)^2 - \frac{E}{\delta t}$$
It seems like if we take the limit as $\delta t \rightarrow 0$, the RHS will explode, but that is not the case. If we consider the way $B, D, E$ are defined, we'll realize that they can be represented by some fixed constant rate of changes $b, d, e$ multiplied with the time interval $\delta t$, e.g. $B = b \delta t$. Of course the meaning changes slightly, but significant between $B$ and $b$. The equation that we end up with using this changes of variable is:
$$\frac{N(t+\delta t) - N(t)}{\delta t} = b N(t) - d N(t)^2 - e$$
Taking the limit $\delta t \rightarrow 0$ with assumption on smoothness of $N$, then:
$$N'(t) = b N(t)\left[ 1 - d N(t)\right] - e$$
