Understanding the Limit Rule for sin(3t)cos(5t) [SOLVED]

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Homework Help Overview

The discussion revolves around understanding the limit of the expression sin(3t)cos(5t)/sin(5t) and the reasoning behind breaking it into three separate limits. Participants are exploring the limit rules and theorems applicable to trigonometric functions.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants question how the original expression was split into three limits and the rationale behind this breakdown. There is discussion about the placement of cos(5t) in the expression and its implications for the limit calculation.

Discussion Status

Some participants have offered insights into the mathematical correctness of the breakdown, while others are still seeking clarity on specific aspects of the limit process. There is acknowledgment of a potential mistake in the original solution regarding the placement of cos(5t>.

Contextual Notes

Participants are working within the constraints of a homework problem and are referencing a specific solution document. The discussion reflects uncertainty about the application of limit rules to trigonometric functions.

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[SOLVED] Question on a limit rule

Homework Statement


http://www.math.yorku.ca/Who/Faculty/Kochman/M1300/solutions/solF06/SFE.pdf
Question 3. My question is (Using t as theta, as I'm too lazy to use TeX), when going from the limit of sin(3t)cos(5t)/sin(5t), the person then breaks up this limit into 3 separate limits (All multiplied by each other), then reduces those 3 limits down to 3/5, 1, and 1.

Where did those 3 limits come from? What is the rule or theorem that I'm missing here?
I get how they break down into 3/5, 1, and 1, but I don't see how he split the first into those 3.

Homework Equations


None.


The Attempt at a Solution


Google? :P
 
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They've just multiplied the top and bottom of the expression by 3, 5 and theta. That is, they have multiplied the expression by 1, three times.
 
First of all, there's a mistake in the solution; in the first of the three limits, the cos(5t) should be in the numerator, not the denomator.

Correcting this, the product of the three expressions (before you take the limit), equals the original expression, so this breakup is mathematically correct.

Why choose this particular form? Because we know sin(x)/x -> 1 as x->0, so it's useful to put in a factor of 1/x for each sin(x) (in the numerator or denominator). The leftover stuff then turns out be a number times cos(x), and the limit of this is easy as well. So the general idea was to write the original expression as a product of expressions whose limits are well known.
 
cristo said:
They've just multiplied the top and bottom of the expression by 3, 5 and theta. That is, they have multiplied the expression by 1, three times.
At first I thought I got it, but then something else threw me off.

Without multiplying by 1, sin(3t)cos(5t)/sin(5t) would break into:
Limit of [ sin(3t) / 1 ]
Limit of [ cos(5t) / 1 ]
Limit of [ 1 / sin(5t) ]

However their three limits have one sin on the top, one sin on the bottom, and one cos on the bottom.

How did their cos get to the bottom of the fraction?
 
Goldenwind said:
How did their cos get to the bottom of the fraction?
Yea, sorry, I didn't notice that; see the above post by Avodyne.
 
Good to know. Thank-you for your help :)
/solved
 

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