Easy Calculus Limit Problem. 10 points.?

Question

lim (x approaches 0) (sin3t)/(2t)

Explain step by step please.

I'm trying to understand the process of this. I haven't learned the L' Hospital theorem, but learned (sinx)/x = 1. Can someone please explain how this applies to the above problem, and walk me through the problem?

Answer 1

Well, to get the sin(t)/t we want, the argument of sin must be simply t, right? So we do a substitution, stating that u=3t. Now we can write this as:

sin(u)/(2/3*u)

To pull that 2/3 out (since we need the denominator to be just u), change the "division by a fraction" into "multiplication by its reciprocal". The reciprocal of 2/3 is 3/2, so we end up with:

3/2 * sin(u)/u

We know that the limit of 3/2 is ALWAYS 3/2, no matter what t is approaching. We know that as t->0, u->0 (because u=3t, and 3t->0 as t->0). And since we know that as u->0, sin(u)/u->1, we end up with:

(3/2)*1=3/2

-IMP ;) :)

*EDIT* @Bondman: He said he didn't want to use L'Hospital's Rule

Answer 2

Ugh for "sam's" reasoning to help you...just think of letting x = 3t....sin t / [2t / 3 ]---> {sin t / t} { 1 / [2/3]} = [3/2] {sin t / t }....or multiply top and bottom by 3...3 sin (3t) / [2(3t)] = [3/2] {sin (3t) / (3t) }= [3/2] sin x / x....likewise for something like [sin 5t / sin 3t] = {sin 5t / 5t} / {sin 3t / 3t} { 5t / 3t}----> (1)(1) 5/3

Answer 3

sin(3t)/(2t). Use l'hopital's rule since the limit is in the form 0/0.

So differentiate the top and bottom and divide.

3/2 * sin(3t)

lim t --> 0 of 3/2*sin(3t) = 0

If you meant sin^3(t)/(2t) then do the same thing:

3(sin(t))^2(cos(t))/2 = 0 as well since sin(0) = 0

Answer 4

ok so sin3t/2t you can pull our sint/t out and make it a 1 then divide 3 by 2 and multiply it the 1 you pulled out

Sin(3t)

2t

1(3)

  (2)

1(1.5)

1.5

so when the lim (x approching 0) the lim is 1.5

hope it helps