Limit of sin(3x)/sin(5x) as x→0

  • Thread starter Thread starter PirateFan308
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary

Homework Help Overview

The problem involves finding the limit of the function \(\frac{\sin(3x)}{\sin(5x)}\) as \(x\) approaches 0, which falls under the subject area of calculus, specifically limits and trigonometric functions.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how to prove the limit they calculated using a calculator, expressing confusion about the continuity of the sine function and the implications of division by zero.
  • Some participants suggest manipulating the expression by multiplying by appropriate forms of 1 to facilitate the limit evaluation, referencing the limit property of \(\frac{\sin u}{u}\) as \(u\) approaches 0.

Discussion Status

The discussion includes attempts to clarify the original poster's understanding of the limit and the application of trigonometric properties. Guidance has been offered regarding a potential approach to the problem, though there is no explicit consensus on the method or outcome yet.

Contextual Notes

The original poster expresses uncertainty about the validity of their initial calculation and the continuity of the sine function in this context. There is also a mention of a similar thread, suggesting a potential pattern in the types of questions being asked.

PirateFan308
Messages
91
Reaction score
0

Homework Statement


Find [tex]\lim_{x\rightarrow 0} {\frac{\sin(3x)}{\sin(5x)}}[/tex]


The Attempt at a Solution


I know that the limit equals 0.6 (by typing it into my calculator), but I have no idea how to prove this, or even where to start. I know that sin is continuous, so I theoretically should be able to just plug it in, but obviously this doesn't work because it isn't divisible by 0.
 
Physics news on Phys.org
Try multiplying by 3x/(3x) and 5x/(5x), and placing the numerators and denominators strategically. The basic idea is that [itex]\lim_{u \to 0} \frac{sin u}{u} = 1[/itex]
 
Thank you! I can't believe I didn't think of that answer - that helped me figure out the later questions as well.
 

Similar threads

Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 54 ·
2
Replies
54
Views
5K