Step 1 :Given the rate of growth of the bacteria population as \(20 e^{\frac{t}{4}}\) per unit of time \(t\), we can integrate this rate function to find the function that describes the population size at any given time.
Step 2 :The integral of \(20 e^{\frac{t}{4}}\) is \(80 e^{\frac{t}{4}} + C\), where \(C\) is the constant of integration.
Step 3 :We know that at time \(t=0\), the population size was 2300. Substituting these values into the equation, we get \(2300 = 80 + C\). Solving for \(C\), we find that \(C = 2220\).
Step 4 :Therefore, the function that describes the population size at any given time is \(80 e^{\frac{t}{4}} + 2220\).
Step 5 :Substituting \(t=4\) into this equation, we find that the population size at \(t=4\) is \(80 e + 2220\).
Step 6 :For part (b), we are asked to find the limit of the function \(\frac{\ln (1+x)}{x}\) as \(x\) approaches 0. Using L'Hopital's rule, we find that this limit is \(\frac{1}{1+0} = 1\).
Step 7 :\(\boxed{\text{Final Answer: }}\) (a) The number of bacteria present at \(t=4\) is \(\boxed{80 e + 2220}\). (b) The limit as \(x\) approaches 0 of the function \(\frac{\ln (1+x)}{x}\) is \(\boxed{1}\).