Find the limit. limt →∞ (t-t √(t))/(2 t^3 / 2+3 t-5) |...

Name: Problem 22, Chapter 1: Essential Calculus Early Transcendentals
Uploaded: 2020-05-28T23:23:35-08:00
Duration: 3 min 18 s
Channel: Yaw Asomani
Description: Find the limit. limt →∞ (t-t √(t))/(2 t^3 / 2+3 t-5)

Key Concepts

Limits

Limits are a fundamental concept in calculus used to determine the behavior of a function as its input approaches a certain value, including infinity. Understanding limits is essential for analyzing continuity, convergence, and the behavior of functions near points of interest or at boundaries.

Dominant Term Analysis

When evaluating limits, especially for functions involving expressions with power terms, identifying the dominant terms (those that grow the fastest) in the numerator and denominator is crucial. This approach allows one to simplify the expression by focusing on the terms that have the greatest impact on the function's behavior as the variable approaches infinity.

Asymptotic Behavior

Asymptotic behavior describes how a function behaves as its input grows infinitely large or approaches a particular point. This concept simplifies the analysis of complex functions by highlighting the leading-order terms, thus making it easier to evaluate limits and understand the long-term trends of the function.

Transcript

00:01 So we have on our plate limit as t goes to infinity, right? as t goes to infinity of t minus t, right, square root of t over t to the power, 2t to the power 3 over 2, right? plus 3t minus 5, right? we have this one on our plate to solve.

00:34 We just have to find what is the highest power of t and then divide.

00:39 But the highest power of t is t to the power 3 over 2.

00:43 So we're going to divide each and every term by t to the power 3 over 2, right? remember, this one here is also t to the power 3 over 2, right? because t times square root of t is the same as t times t to the power 1 half.

01:00 Square root of t is t to the power one half and this is t to the power of one so when you add one plus one half that is also going to be three over two as well right so this one is going to be i'm just going to divide each and every term by the highest power of t which is t to the power three over two right so this is three over two then minus the numerator this one the entire thing here is t to the power three over two right and i'm divided it by t by t to the power 3 over 2 as well, right? and then here you're gonna have two times t to the power 3 over 2, also divided by t to the power 3 over 2, right? and then minus 3 t over t to the power 3 over 2, and then minus, this is plus and then minus 5 over t to the power 3 over 2.

01:52 Right, so you can see that this one here, right? this one here becomes one, right? and then this one here becomes two, right? and then so what is happening? you have limit t approaches infinity.

02:09 Now this is one here.

02:11 So subtract those one from this three over two...

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