Evaluate the limit. limx →0 (sin3 x sin2 x)/(x sin5 x) |...

Name: Problem 43, Chapter 2: Calculus for AP
Uploaded: 2020-02-22T02:46:51-08:00
Duration: 4 min 36 s
Channel: Justin Leonesio
Description: Evaluate the limit. limx →0 (sin3 x sin2 x)/(x sin5 x)

Key Concepts

Standard Trigonometric Limit

A common result in calculus is the limit lim (x ? 0) (sin x)/x = 1, which is derived from the geometry of the circle and is essential for analyzing small angles. This property is fundamental when dealing with trigonometric limits, allowing more complex expressions involving sine functions to be manipulated and simplified by factoring out terms that resemble sin x/x.

Algebraic Manipulation in Limit Evaluation

Algebraic manipulation involves rewriting expressions to apply known limits effectively. In limit problems, especially those with products or quotients of functions, algebraic techniques such as factoring and grouping terms help isolate instances of the standard sine limit, facilitating the evaluation of the overall limit.

Limits

Limits are foundational concepts in calculus that describe the behavior of a function as its input approaches a particular value. They are crucial for understanding continuity, defining derivatives, and evaluating behaviors at boundaries or singular points. In this context, evaluating the limit of a trigonometric ratio as x approaches 0 involves determining its precise value based on the behavior of its components near that point.

Transcript

00:01 All right, for this example, we're going to attempt to solve for the limit as x approach is zero for this expression.

00:06 And if you look at using the substitution method initially here, you'll be able to tell pretty easily that you'll get a zero on top and a zero on the bottom.

00:14 So with an indeterminate form like that, we're going to need to use something different.

00:17 And in particular, they give us some tools here to be able to use it from the section.

00:21 And theorem two, rather, that they give us this limit was involved.

00:26 This is a tool that they're providing for us.

00:28 And in previous examples, we've looked at creating kind of a brand new tool from this old one here.

00:34 Notice that if sign of x over x is approaching 1 as the x values approach 0, well, then the reciprocal should also be true.

00:41 You know, is x approach 0? if you flip this upside down, you should also be approaching 1 because that means that roughly the top and the bottom are approaching the same result, leading us towards 1 overall.

00:51 So this should hold 2, and thus both of these tools can be utilized within the problem if we need it.

00:57 So to start actually working on this, it might be beneficial if we break it up into multiple pieces all multiply together.

01:03 So we'll start with the limit as x approaches zero still.

01:07 But we're going to start kind of forming some of these if at all possible.

01:11 Now a good way to start off here is let's just keep this sign of 3x here as the numerator of 1 fraction.

01:19 And to help out that expression a little bit for the denominator we're going to need a 3x on the bottom, right? in order for this to work out, we need the expression here within the sign to also appear in the denominator or vice versa.

01:31 So we're going to need a 3x eventually over here underneath the sign of 3x.

01:35 So let's just take this x that we already have and join it up on this side here.

01:40 We'll at least get a head start on that.

01:42 Okay.

01:43 Now for the other parts.

01:44 Notice we have really just the sign of 5x on the bottom and the sign of 2x on the top.

01:48 So let's start separating those into their own fractions.

01:51 I'll put the sign of 5x here in the denominator for this first fraction.

01:56 And then for the second one we'll put the sign of 2x over here now each of these pieces is going to have something else that we're going to create in their denominator or numerator respectively so i'm just going to leave them blank for now picture them as ones for the moment if you'd like okay well remember we have just an x over here we're going to need a 3x down there in order to make this works so what i'm going to do is multiply the denominator by 3x by 3 rather to create the 3x but don't forget if you do it there you're going to have to also do it to the numerator at some point.

02:27 So i'm going to put it out back just to remind us that we did multiply by three at some point.

02:33 Okay...

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