Finding the Limit of Trig Functions: 2x/sin3x, x->0

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The limit of 2x/sin(3x) as x approaches 0 can be evaluated using the known limit lim sin(x)/x = 1. By rewriting the expression as (2/3)(3x/sin(3x)), the limit simplifies to (2/3)(1) as x approaches 0, resulting in a final answer of 2/3. The discussion clarifies that while lim 2/3 sin(x)/x equals zero at x=0, the correct approach for 2x/sin(3x) leads to a non-zero limit. The solution is confirmed to be correct.
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Homework Statement



lim 2x/(sin3x)
x-> 0

Homework Equations



lim sinx/x = 1
x->0

The Attempt at a Solution



is it correct to say the following:

lim 2/3 (sinx/x)
x-> 0

lim 2/3 (1)
x-> 0

Answer: lim 2x/sin3x = 2/3
x-> 0

Because it's on the book:
cos 2x(2)/3 = 2/3
x->0

So I want to know if I solve it the way I did before, would it be correct?
 
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domyy said:

Homework Statement



lim 2x/(sin3x)
x-> 0


Homework Equations



lim sinx/x = 1
x->0

The Attempt at a Solution



is it correct to say the following:

lim 2/3 (sinx/x)
x-> 0

lim 2/3 (1)
x-> 0

Answer: lim 2x/sin3x = 2/3
x-> 0

Because it's on the book:
cos 2x(2)/3 = 2/3
x->0

So I want to know if I solve it the way I did before, would it be correct?

lim 2/3 sin(x)/x is zero at x=0, but 2x/sin(3x) is not the same as (2/3)sin(x)/x.

Rewrite 2x/sin(3x) as 2/3 (3x)/sin(3x).

ehild
 
Thanks! Got it! =)
 

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