Evaluate the following limit by first recognizing the sum as a Riemann sum for a function defined on [0, 1]:
lim 10/n ( (1/n)^1/2 + (2/n)^1/2 + (3/n)^1/2 +.....+ (n/n)^1/2 )
n----> infinity
To evaluate the given limit, we need to recognize the sum as a Riemann sum for a function defined on [0, 1].
First, let's rewrite the sum in a more convenient form:
(1/n)^1/2 + (2/n)^1/2 + (3/n)^1/2 + ... + (n/n)^1/2
To do this, we can rewrite each term as (k/n)^1/2, where k ranges from 1 to n.
Now, let's consider a function f(x) = x^1/2 defined on the interval [0, 1]. We can see that the sum above represents a Riemann sum for f(x) using n subdivisions.
If we divide the interval [0, 1] into n equal subdivisions, each subdivision will have a width of Δx = 1/n. Therefore, the height of each rectangle in the Riemann sum will be f(x_i) = f(k/n) = (k/n)^1/2, where k is the index of each subdivision.
The Riemann sum, S_n, can be calculated by summing the areas of each rectangle:
S_n = (Δx) * [f(x_1) + f(x_2) + ... + f(x_n)]
= (1/n) * [(1/n)^1/2 + (2/n)^1/2 + ... + (n/n)^1/2]
The given limit is the limit of the Riemann sum as n approaches infinity:
lim(n→∞) {(1/n) * [(1/n)^1/2 + (2/n)^1/2 + ... + (n/n)^1/2]}.
To evaluate this limit, we can rewrite it as an integral of f(x) over the interval [0, 1]:
lim(n→∞) {(1/n) * [(1/n)^1/2 + (2/n)^1/2 + ... + (n/n)^1/2]}
= lim(n→∞) {(1/n) * [f(1/n) + f(2/n) + ... + f(n/n)]}
= lim(n→∞) ∑[(1/n) * f(k/n)], where k ranges from 1 to n
= ∫[0,1] f(x) dx.
Now, the limit is equivalent to the integral of f(x) from 0 to 1.
To evaluate this integral, we can use the power rule for integration:
∫ x^1/2 dx = (2/3)x^(3/2).
Evaluating the integral:
∫[0,1] f(x) dx = ∫[0,1] x^1/2 dx = [(2/3)x^(3/2)] evaluated from 0 to 1.
Substituting the limits into the equation:
= [(2/3)(1)^(3/2)] - [(2/3)(0)^(3/2)]
= (2/3)(1) - (2/3)(0)
= 2/3.
Therefore, the limit as n approaches infinity is 2/3.