What is the following limit?
lim as n goes to infinity of (pi/n) (sin(pi/n) + sin(2pi/n) + sin(3pi/n) +...+ sin(npi/n)) =
I.) lim as n goes to infinity sigma (n and k=1) of pi/n sin(kpi/n)
II.) Definite integral from 0 to pi of sin(x)dx
III.) 2
A.) I only
B.) II only
C.) III only
D.) II and III only
E.) I, II, and III
I am really not getting much of this at all, I think I have it narrowed down to A, D, or E.
To find the limit of the given expression, let's break down each option:
I.) lim as n approaches infinity of Σ(k=1 to n) of (π/n)sin(kπ/n):
To solve this option, we can rewrite the expression as a Riemann sum. The Riemann sum approximates the definite integral of a function over an interval by dividing the interval into subintervals and summing the values of the function over these subintervals. In this case, we have Σ(k=1 to n) of (π/n)sin(kπ/n), which represents the Riemann sum.
II.) Definite integral from 0 to π of sin(x)dx:
The integral of sin(x) from 0 to π can be evaluated using standard integration techniques. The integral of sin(x) is -cos(x), and evaluating it from 0 to π gives -cos(π) - (-cos(0)). Simplifying further, we have -(-1) - (-1) = 1 + 1 = 2.
III.) 2:
The constant 2 is a possible limit value.
Comparing the options A, D, and E:
Since option II evaluates to 2 and option III is also 2, the correct answer is D: II and III only.
Therefore, the limit as n goes to infinity of the given expression is 2.
To find the limit, we can write the given expression as a Riemann sum. Let's break it down step-by-step:
1. Rewrite the expression in sigma notation:
lim as n goes to infinity of (pi/n) (sin(pi/n) + sin(2pi/n) + sin(3pi/n) +...+ sin(npi/n)) = lim as n goes to infinity of sigma (k=1 to n) (pi/n) sin(kpi/n)
2. Recognize that the Riemann sum of sin(x) over the interval [0, pi] can be represented as:
Definite integral from 0 to pi of sin(x) dx = lim as n goes to infinity of sigma (k=1 to n) (pi/n) sin(kpi/n)
3. Compare the two expressions:
We can see that the given limit is equivalent to the definite integral from 0 to pi of sin(x) dx. Therefore, the expression simplifies to:
lim as n goes to infinity of (pi/n) (sin(pi/n) + sin(2pi/n) + sin(3pi/n) +...+ sin(npi/n)) = Definite integral from 0 to pi of sin(x) dx
4. Evaluate the definite integral:
The definite integral of sin(x) from 0 to pi is equal to 2.
Therefore, the correct answer is:
C.) III only