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

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SUMMARY

The limit of sin(4x)/3x as x approaches 0 can be evaluated using the Squeeze Theorem. The transformation sin(4x)/(4x) approaches 1 as x approaches 0, which is established by the limit u->0 of sin(u)/u=1, where u=4x. Thus, the limit lim (sin(4x)/3x) as x approaches 0 equals 4/3.

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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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