3-6 Find the limit. limt →0(e^-3 t 𝐢+(t^2)/(sin^2 t) 𝐣+cos2 t 𝐤)

Name: Problem 3, Chapter 13: Calculus Early Transcendentals
Uploaded: 2020-10-13T16:45:53-08:00
Duration: 4 min 9 s
Channel: Bobby Barnes
Description: 3-6 Find the limit. limt →0(e^-3 t 𝐢+(t^2)/(sin^2 t) 𝐣+cos2 t 𝐤)

3-6 Find the limit. limt →0(e^-3 t 𝐢+(t^2)/(sin^2 t) 𝐣+cos2...

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Key Concepts

Exponential Function Behavior

Exponential functions, such as e^(at), are continuous and well-behaved at points like zero. Their limits are easily obtained by plugging the approaching value into the function, which is especially useful in limit problems.

Trigonometric Function Limits

Trigonometric functions have standard limits that are often used to simplify expressions. For example, recognizing limits like sin(t)/t equal to 1 when t approaches zero helps in handling more complex expressions involving trigonometric functions.

Vector-Valued Function Limits

This concept involves understanding how to evaluate the limit of a function that outputs a vector instead of a scalar. If each component function of the vector has a limit as the variable approaches a certain value, then the limit of the vector function is simply the vector composed of these individual limits.

Component-wise Limit Computation

The limit of a vector-valued function can be found by computing the limit for each individual component. This property simplifies the process because you can apply the standard limit laws to each component independently.

Transcript

00:01 So they want us to find the limit as t approaches zero of this parametric equation here.

00:07 So what we're going to do is think about this just like we did back in calculus 1, where we would distribute this limit across the addition.

00:20 So i'll just go ahead and write out that stuff first.

00:23 So this is going to be first the limit as t approaches 0 of e to 2.

00:30 To the negative 3 t plus the limit as t approaches 0 of t squared over sine squared t j and then plus the limit as t approaches 0 of cosine to k so this first limit well we can just go ahead and plug this in because we know e to the negative 3t is just continuous for all values of t.

01:13 So that would be e to the negative 3 times 0 or just 1.

01:19 So that would be 1 times i.

01:25 And then plus.

01:28 So we plug 0 in so we get 0 squared.

01:32 Sign of 0 is also 0.

01:34 So that would be 0 over 0.

01:36 So we need to apply lopetals for this.

01:38 So actually let's come back to this one.

01:40 We'll do this off on the side.

01:43 I'll do that over here.

01:46 And then let's see about this last one.

01:49 So cosine we know is continuous for all values of d.