3-6 Find the limit. limt →1((t^2-t)/(t-1) 𝐢+√(t+8) 𝐣+(sinπt)/(lnt) 𝐤)

Name: Problem 4, Chapter 13: Calculus Early Transcendentals
Uploaded: 2020-10-13T16:52:46-08:00
Duration: 4 min 34 s
Channel: Bobby Barnes
Description: 3-6 Find the limit. limt →1((t^2-t)/(t-1) 𝐢+√(t+8) 𝐣+(sinπt)/(lnt) 𝐤)

3-6 Find the limit. limt →1((t^2-t)/(t-1) 𝐢+√(t+8)...

Key Concepts

Vector-Valued Functions

A vector-valued function is a function whose output is a vector rather than a scalar. When evaluating limits for vector-valued functions, each component (or coordinate) of the vector is considered separately. The overall limit exists if and only if the limits of all individual components exist.

Component-Wise Limit

The concept of a component?wise limit indicates that to determine the limit of a vector-valued function, one computes the limit of each of its individual components independently. This approach simplifies the problem, as each coordinate can often be treated using standard single-variable calculus techniques.

Algebraic Simplification

Algebraic simplification involves manipulating an expression into a simpler form that is easier to work with. This may include factoring, canceling common factors, or rewriting expressions to remove indeterminate forms such as 0/0. Simplification is an essential step to reveal the underlying behavior of a function near a point of interest.

L'Hôpital's Rule

L'Hôpital's Rule provides a method for evaluating limits that initially present forms like 0/0 or ?/?. By differentiating the numerator and denominator and then taking the limit, one can often determine the limit of an expression that otherwise appears indeterminate. This technique is particularly useful when simpler algebraic manipulation is not sufficient.

Continuity

Continuity refers to the property of a function where small changes in the input result in small changes in the output. In the context of limits, if a function is continuous at a point, then the limit as the variable approaches that point is simply the value of the function at that point. This property can expedite the limit evaluation process for well-behaved functions.

Transcript

00:01 So they want us to find the limit of this vector -valued function as t approaches 1.

00:07 So what we can do is think about this just like we did in calculus 1, where we distribute our limit across the plus.

00:16 So let's go ahead and just do that.

00:18 So this is going to be equal to the limit as t approaches 1 of t squared minus t over t minus 1, i and then plus the limit as t approaches one of the square root of t plus eight j plus the limit as t approaches one of sine of pi t over natural log of t times k and now we can just apply this limit individually and so that first one we won't be able to evaluate wait directly because notice if we plug one in, we're going to get 0 over 0.

01:07 So we'll have to do a little bit of algebra to simplify this down.

01:10 So i'll just write this off on the side over here.

01:14 So as t approaches 1 of t squared minus t over t minus 1.

01:21 And then, so this middle one here, we know root t plus 8 is going to be continuous at 1 so we could just plug it in directly.

01:31 So that would give us 1 plus 8, 9, root 9 is 6.

01:35 3.

01:35 So that middle term is going to be plus 3j.

01:39 And then this last one over here, so we plug in 1, sine of pi is 0, natural log of 1 is 0.

01:47 So we'll need to do something with this expression also.

01:49 So i'll do that off on the side over here as well.

01:54 Or actually, instead of just doing that, i'll put it over here.

01:59 And then i'll just draw a line to separate this.

02:04 So technically in both cases, since we have 0 over 0 we could use lopatals, but you might recall from calculus 1 that when we solve something like this, we could try to factor it.

02:16 And notice that that numerator actually factors to t -t minus 1.

02:21 So the t -minus 1s cancel out with each other.

02:24 And we're just going to be left with the limit as t approaches 1 of t...

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