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

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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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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