Limit of sin4x/3x as x approaches 0: Squeeze Theorem

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Homework Help Overview

The discussion revolves around finding the limit of the expression sin(4x)/(3x) as x approaches 0, specifically using the Squeeze Theorem. The subject area pertains to limits and trigonometric functions.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the algebraic manipulation of the limit expression and question the reasoning behind the limit of sin(4x)/(4x) approaching 1. There is a focus on understanding the foundational limit of sin(u)/u as u approaches 0.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the limit properties and confirming the validity of the limit of sin(u)/u. There is no explicit consensus yet, but some guidance has been offered regarding the relationship between the limits involved.

Contextual Notes

Participants reference the assumption that the limit of sin(u)/u as u approaches 0 is known, suggesting that this is a foundational concept expected in the context of the problem.

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Homework Statement


Find lim sin4x/3x
x -> 0


Homework Equations





The Attempt at a Solution


I did some algebraic massaging and got

sin4/3x = 4/4 * sin4x/3x = 4/3 * sin4x/4x
but I don't understand how sin4x/4x becomes 1 like my textbook says so.
 
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sin(4x)/(4x) isn't 1. The limit as x->0 of sin(4x)/(4x) is 1. Have you proved limit u->0 of sin(u)/u=1? Then just put 4x=u.
 
Last edited:
Dick said:
sin(4x)/(4x) isn't 1. The limit as x->0 of sin(4x)/(4x) is 1. Have you proved limit u->0 of sin(u)/u=1? Then just put 4x=u.

Yeah you're right.So, It's true that limit u->0 of sin(u)/u=1?
 
Didn't you prove limit u->0 sin(u)/u=1. I doubt they would have given you this problem if they hadn't.
 

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