Hi all. My problem is to take the limit as t approaches 0 of sin(3t) divided by sin(8t). The first step was to split the fraction. So now I have the limit as t approaches 0 of (sin3t))/1 times 1/(sin8t)). I know that some kind of way I should multiply both fractions by t so I can use the fact (limit as theta approaches 0 (sin theta/theta)=1. When I multiplied the fractions by (t), I think that I should get (sin(3t^2))/t times (t/(sin8t^2)). However, I am wrong and I honestly don't understand why I am wrong. Please help.
Limits involving trig functions
- Thread starter mikagurl
- Start date
You have the right idea but just need to go a step more. We have, as you mentioned,
sin(x)/x -> 1
but sin(3x)/x is not of that form. However sin(3x)/(3x) is of that form and
sin(3x)/x = 3 * [sin(3x)/(3x)]
Steven G
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- Dec 30, 2014
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I am a bit more concerned that you think that tsin(3t) = sin (3t^2). This would mean that you think that sin(5y) = 5 sin y.
Now think of this. According to this logic then sin(75) = sin (75*1) = 75 (sin 1). So all we need to know is what sin(1) equals. Make believe it equals the special number k. Then sin (75) =75k, sin (2/3)=(2/3)k. If this is correct then evaluating sines would be easier then we have been taught which makes no sense. Also if the argument (what you are taking sin of) is large enough then the sin of a large number would be more than 1, which is not possible.
Here is the way to solve your problem: sin (ax)/ sin(bx) = [sin (ax)/ax]/[sin (bx)/bx] * ax/bx = [sin (ax)/ax]/[sin (bx)/bx] * a/b. Now as x->0, this product equals [1]/[1] * a/b = a/b. So you can read off the answer from the start!
Jomo
Now think of this. According to this logic then sin(75) = sin (75*1) = 75 (sin 1). So all we need to know is what sin(1) equals. Make believe it equals the special number k. Then sin (75) =75k, sin (2/3)=(2/3)k. If this is correct then evaluating sines would be easier then we have been taught which makes no sense. Also if the argument (what you are taking sin of) is large enough then the sin of a large number would be more than 1, which is not possible.
Here is the way to solve your problem: sin (ax)/ sin(bx) = [sin (ax)/ax]/[sin (bx)/bx] * ax/bx = [sin (ax)/ax]/[sin (bx)/bx] * a/b. Now as x->0, this product equals [1]/[1] * a/b = a/b. So you can read off the answer from the start!
Jomo
Jomo, what you just explained makes more sense to me now. It think I was thinking about polynomials. For an example if I have x^3*(x) it would equal x^4 because we would add the exponents. I think that is what I was trying t do here. Thanks for you help.