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Squeeze theorem question

  • Thread starter appplejack
  • Start date
1. The problem statement, all variables and given/known data
Find lim sin4x/3x
x -> 0


2. Relevant equations



3. 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 text book says so.
 

Dick

Science Advisor
Homework Helper
26,255
618
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:
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?
 

Dick

Science Advisor
Homework Helper
26,255
618
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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