The interval [0,5] is partitioned into n equal subintervals and a number xi is defined to be the left hand endpoint of the ith subinterval for every i. Then
n-1
limit of n going to infinity Σ (2xi-5)/n = what?
i=0
To determine the limit of the sum as n approaches infinity, we can rewrite the given expression:
Σ (2xi - 5)/n
First, let's express xi in terms of i and n. Since the interval [0,5] is partitioned into n equal subintervals, the length of each subinterval is:
Δx = (5 - 0)/n = 5/n
Therefore, the left-hand endpoint of the ith subinterval (xi) can be calculated as:
xi = (i * Δx) + 0
= (i * 5/n)
Now, let's substitute xi back into the original expression:
Σ (2xi - 5)/n
= Σ (2 * (i * 5/n) - 5)/n
= Σ (10i/n - 5)/n
= (1/n) * Σ (10i - 5n)/n
Now, let's look at the numerator, which is a sum of arithmetic series: Σ (10i - 5n).
Using the formula for the sum of an arithmetic series, the sum of the terms from i = 0 to n-1 is given by:
Σ (10i - 5n) = (n/2) * (10(0) + 10(n-1)) - 5n
= (n/2) * (10n - 10) - 5n
= (10n^2 - 10n - 10n + 10) / 2 - 5n
= (10n^2 - 20n + 10) / 2 - 5n
= (5n^2 - 10n + 5) / 2 - 5n
= (5/2)n^2 - (20/2)n + (5/2) - 5n
= (5/2)n^2 - 10n + (5/2) - 5n
= (5/2)n^2 - 15n + (5/2)
Now, substitute this back into the original expression:
(1/n) * Σ (10i - 5n)/n
= (1/n) * [(5/2)n^2 - 15n + (5/2)]
= (1/n) * [(5/2)n^2/n - 15n/n + (5/2)/n]
= (5/2n^2) - (15/n) + (5/2n)
Next, let's take the limit as n approaches infinity:
lim (n → ∞) (5/2n^2) - (15/n) + (5/2n)
As n approaches infinity, the terms with n in the denominator become negligible compared to the term with n^2 in the denominator. Therefore, the limit simplifies to:
lim (n → ∞) (5/2n^2)
The limit of this expression is 0.
Therefore, the limit of the given sum as n approaches infinity is 0.