MATH 131
Final Exam
June 29,2023
5. (a) (6 points) The number of bacteria in a culture is growing at a rate of $20 e^{\frac{t}{4}}$ per unit of time $t$. At time $t=0$, the number of bacteria present was 2300 . Find the number present at $t=4$.
(b) (4 points) Find the limit
\[
\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}
\]

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 \(\boxed{1}\).