- One batch of walnut muffins uses 1(1)/(3) cups of walnuts. How many cups of walnuts are needed to make 3(3)/(4) batches of muffins?
- Express the following limit in terms of the number e. limk?8?(1+9k1?)5k
The correct answer and explanation is:
1. How many cups of walnuts are needed to make 3(3)/(4) batches of muffins?
To solve this, we need to determine how many cups of walnuts are required for 3(3/4) batches of walnut muffins, given that 1(1/3) cups of walnuts are used for one batch.
Step 1: Convert the mixed fractions to improper fractions.
- 1(1/3) = (3/3) + (1/3) = 4/3 cups of walnuts per batch.
- 3(3/4) = (12/4) + (3/4) = 15/4 batches of muffins.
Step 2: Multiply the number of batches by the cups of walnuts per batch.
Now, multiply 4/3 cups (for one batch) by 15/4 batches: 43×154=4×153×4=6012=5 cups of walnuts.\frac{4}{3} \times \frac{15}{4} = \frac{4 \times 15}{3 \times 4} = \frac{60}{12} = 5 \text{ cups of walnuts}.
Thus, 5 cups of walnuts are needed for 3(3/4) batches of muffins.
2. Express the following limit in terms of the number e:
The limit is given as: limk→∞(1+9k)5k.\lim_{k \to \infty} \left( 1 + \frac{9}{k} \right)^{5k}.
Step 1: Recognize the standard form for the limit.
This expression resembles the standard form of the exponential limit: limk→∞(1+ak)bk=eab.\lim_{k \to \infty} \left( 1 + \frac{a}{k} \right)^{bk} = e^{ab}.
In this case, a=9a = 9 and b=5b = 5. So, we can directly apply the formula: limk→∞(1+9k)5k=e9×5=e45.\lim_{k \to \infty} \left( 1 + \frac{9}{k} \right)^{5k} = e^{9 \times 5} = e^{45}.
Thus, the limit simplifies to e45e^{45}.
Explanation:
The expression (1+9k)5k\left( 1 + \frac{9}{k} \right)^{5k} is a typical example of a limit that approaches an exponential function as kk becomes large. As kk increases, 9k\frac{9}{k} becomes smaller, and the expression begins to resemble the well-known limit definition of the number ee, which is approximately: e=limn→∞(1+1n)n.e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n.
In our case, the factor inside the parentheses is 9k\frac{9}{k}, and the exponent is 5k5k. Therefore, using the general rule for exponential limits, we find that the expression converges to e45e^{45}. This result emphasizes the power of the constant ee in modeling exponential growth or decay, and it is often encountered in calculus and applications involving continuous compounding, such as in finance and population growth models.
In conclusion, the limit is e45e^{45}, which provides an elegant way to express growth over time in terms of the natural exponential function.