Why Does (sin(2x))/x Equal 2 as x Approaches 0?

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Homework Help Overview

The discussion revolves around the limit of the expression (sin(2x))/x as x approaches 0, with participants exploring why this limit equals 2. The subject area is calculus, specifically limits involving trigonometric functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the limit of (sin(2x))/x and its relationship to known limits involving sine functions. Some express uncertainty about the reasoning behind the limit equating to 2, while others suggest using identities or adjustments to the expression.

Discussion Status

The discussion is active, with various approaches being considered. Some participants have offered guidance on potential methods to evaluate the limit, while others are questioning the completeness of the problem statement and exploring different interpretations.

Contextual Notes

There is mention of the need for a complete problem statement, as well as the context of continuity for a piecewise function involving the limit in question. Participants are also reflecting on known limits and their implications for the current problem.

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Homework Statement


I have the equation (sin(2x))/x = ?


The Attempt at a Solution



I know that the answer to this is 2, but I am not sure why (sin(2x))/x = 2

Can someone explain?

Thanks
 
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hahaha158 said:

Homework Statement


I have the equation (sin(2x))/x = ?


The Attempt at a Solution



I know that the answer to this is 2, but I am not sure why (sin(2x))/x = 2

Can someone explain?

(sin(2x))/x ≠ 2, so perhaps you are leaving something out of the problem. What is the complete problem statement?
 
Mark44 said:
(sin(2x))/x ≠ 2, so perhaps you are leaving something out of the problem. What is the complete problem statement?

Find the value of the constant a for which the function below is continuous everywhere. Fully
explain your reasoning.

... a+x2 while x≤0
f(x) = {
... (sin(2x))/x while x>0
 
The problem is really asking about limits, namely
$$ \lim_{x \to 0}\frac{sin(2x)}{x}$$

Do you know any other limits that involve trig functions?
 
Mark44 said:
The problem is really asking about limits, namely
$$ \lim_{x \to 0}\frac{sin(2x)}{x}$$

Do you know any other limits that involve trig functions?

I'm not sure what you mean by your question.

I know that $$ \lim_{x \to 0}\frac{sin(x)}{x}$$ is equal to 1

I also know that the limit as x approaches 0 from the negative and the limit at x=0 are both just equal to a.

So this means that a just equals $$ \lim_{x \to 0}\frac{sin(2x)}{x}$$

I know the answer to the question is 2 so that means $$ \lim_{x \to 0}\frac{sin(2x)}{x}$$ must equal 2 but I am not sure how to do it.
 
There are at least a couple of ways to go.
1) Double angle identity for sine
2) Adjust things so that you have sin(2x)/(2x) times some other stuff.
 
Mark44 said:
There are at least a couple of ways to go.
1) Double angle identity for sine
2) Adjust things so that you have sin(2x)/(2x) times some other stuff.

For 2) do you mean like

(sin(2x))/2x)*2

=so you get 1*2

=2?

Would that work?
 
(sin(2x))/ x = (sin2(1))/ 1
= sin(2)(1)/1
= sin(2)
= 0.0349
 
5ymmetrica1 said:
(sin(2x))/ x = (sin2(1))/ 1
= sin(2)(1)/1
= sin(2)
= 0.0349

That's the limit as x goes to 1, not 0.

hahaha158 said:
For 2) do you mean like

(sin(2x))/2x)*2

=so you get 1*2

=2?

Would that work?
Yep, exactly! This would be the simplest way to do it, so if you ever get a question like

[tex]\lim_{x\to 0}\frac{\sin(ax)}{b}[/tex] then this is equivalent to
[tex]\lim_{x\to 0}\frac{a}{b}\cdot\frac{\sin(ax)}{a}=\frac{a}{b}[/tex]
 

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