Problem

Justify (3) for the case $ h < 0 $.

01:02

Problem 76 Hard Difficulty

Evaluate the limit by first recognizing the sum as a Riemann sum for a function defined on $ [0, 1] $.

$ \displaystyle \lim_{n \to \infty} \frac{1}{n} \biggl( \sqrt{\frac{1}{n}} + \sqrt{\frac{2}{n}} + \sqrt{\frac{3}{n}} + \cdots + \sqrt{\frac{n}{n}} \biggr) $

Answer

$$
\frac{2}{3}
$$

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Video Transcript

So they first want us to rewrite this as a Remond some s. So let's go ahead and do that. So this would be the limit as it approaches infinity of one of her. And now that was the thing that's changing is the power in the art of the numerator here. So it's like 123 and just kind of keeps increasing world. The denominator always stays in. So this is gonna be the sum from I is equal to one. Or actually, I guess we would start from zero. Um because I would be the same thing to in of the square root of I over in. And now this. If we were to go ahead and pull this one of her in inside so little limit as an approach to the affinity of the some from I 020 to in off one over in hi over in notice that this is now in the form of If so, if we're on 01 this is like B minus a over in where B is one a zero. So this is Delta X eso There's just one my zero over in which is one or in. So that's our Delta X Here, This is Delta X, and then this function is going to be like X I because it's just going to be our Delta X plus R a Oh, Delta x times I like that. Um, yeah, so you could see how. Okay, well, we have one over in times I and then plus zero so that there is Route X, I and so. But this is really going to say is that this is equal to the integral from zero to one of the square root of X dx. And now we can go ahead and evaluate this using power rule truck, my pin. So remember, we can rewrite this as a one half power. So powerful says, add one to the power, so it's gonna be x 23 house, and then we divide by the new power relapse. Then we evaluate this from 0 to 1. Um, so that would just be two thirds next to the three half evaluated from 01 So we plug in one, so be two thirds times one minus two thirds times zero. So that's just zero. So we end up where that infinite limit ends up being equal to two thirds

Bobby B.

University of North Texas

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Problem

Justify (3) for the case $ h < 0 $.

Problem 76 Hard Difficulty

Evaluate the limit by first recognizing the sum as a Riemann sum for a function defined on $ [0, 1] $.

$ \displaystyle \lim_{n \to \infty} \frac{1}{n} \biggl( \sqrt{\frac{1}{n}} + \sqrt{\frac{2}{n}} + \sqrt{\frac{3}{n}} + \cdots + \sqrt{\frac{n}{n}} \biggr) $

Answer

$$
\frac{2}{3}
$$

More Answers

Related Courses

Calculus: Early Transcendentals

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Video Transcript

Calculus: Early Transcendentals

Anna Marie V.

Heather Z.

Caleb E.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

$$\frac{2}{3}$$

Calculus: Early Transcendentals

Grace H.

Catherine R.

Anna Marie V.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Anna Marie V.

Heather Z.

Caleb E.

Samuel H.

$$
\frac{2}{3}
$$