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

  • Thread starter Thread starter Goldenwind
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary
SUMMARY

The discussion focuses on the limit rule applied to the expression sin(3t)cos(5t)/sin(5t). The key takeaway is the breakdown of this limit into three separate limits, resulting in values of 3/5, 1, and 1. The method involves multiplying the expression by 1 in the form of 3, 5, and theta, allowing the use of the known limit sin(x)/x as x approaches 0. The correction noted in the discussion highlights that cos(5t) should be in the numerator, not the denominator, which is crucial for accurate limit evaluation.

PREREQUISITES
  • Understanding of limit theorems in calculus
  • Familiarity with trigonometric functions and their limits
  • Knowledge of the sin(x)/x limit as x approaches 0
  • Basic algebraic manipulation of fractions
NEXT STEPS
  • Study the application of limit theorems in calculus
  • Learn about trigonometric limits and their properties
  • Explore the concept of multiplying by 1 in limit evaluations
  • Practice solving limits involving products of trigonometric functions
USEFUL FOR

Students studying calculus, particularly those focusing on limits and trigonometric functions, as well as educators looking for clear explanations of limit breakdown techniques.

Goldenwind
Messages
145
Reaction score
0
[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
 
Physics news on Phys.org
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
 

Similar threads

Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 27 ·
Replies
27
Views
4K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K