How do I solve this calculus question?

Question

find the limit as t approaches 0 of this function: sin^2(3t)/(t^2).
In words, the function is sine squared times 3t divided by t squared. Any help with trig limits and derivatives is also helpful.

In the numerator: sin^2 3t<---that is, sine squared times 3t written as sin with a 2 exponent and then 3t.

In the denominator: t squared or t^2

Answer 1

Limt { [ sin^2 (3t) ] / [t^2 }
t-->0

Multiple by 9/9

Limt {9/9 [ sin^2 (3t) ] / [t^2 }
t-->0

Limt {9 [ sin^2 (3t) ] / [9t^2 }
t-->0

9 Limt { [ sin^2 (3t) ] / [3t]^2 }
t-->0

since
Limt sin u /u = 1
u-->0

Then let 3t = u and as t-->0 so u-->0

9 Limt { [ sin (u) ] / [u] }^2
u-->0

9 Limt { [ sin (u) ] / [u] }* { [ sin (u) ] / [u] }
u-->0

9 *1*1 = 9

Answer 2

I believe the function is (sin^2(3t))/(t^2). At t=0, this equation becomes the inderminate expression 0/0. This type of problem is solved using L'hospital's rule in which you take the derivative of the numerator expression and the derivative of the denominator expression separately, then evaluate the ratio of the derivatives at t=0.

i.e. if an expression u(x)/v(x) evaluates to 0/0 (or infinity over infinity) at the limit "a", then the limit value is

d(u)dx / d(v)/dx

evaluated at x = a.

Answer 3

sin^2(0) = 0 and 0^2 = 0.
Plugging in t = 0 yields 0/0 - an indeterminate form. Thus we can use L'Hopital's rule...

The derivative of sin^2(3t) = 2*sin(3t)*cos(3t)*3 ( use the chain rule)
Thus [ sin^2(3t) ]' = 3*sin( 6t )

The derivative of t^2 is 2t

Thus the limit is equivalent to the limit, as t approaches 0, of
[ 3 * sin( 6t ) ] / [ 2t ]

Plugging in 0 again, we see that this too is an indeterminate form. L'Hopital's rule again!

[ 3 * sin( 6t ) ] ' = 18 * cos( 6t )
[ 2t ] ' = 2

Thus the limit is the same as the limit as t approaches 0, of
[18 * cos( 6 t ) ] / 2

This is easily seen to be [ 18 * 1 ] / 2 = 9

Then: lim_{t \to 0} sin^2(3t)/(t^2) = 9


Let's check this a different way!
We know that lim_{x \to 0} sin(x)/x = 1. We can use this to find the limit.

lim_{t \to 0} sin^2(3t)/(t^2)
= lim_{t \to 0} [ sin(3t)/t ]^2
= [ lim_{t \to 0} sin(3t)/t ]^2

Thus we need only analyze lim_{t \to 0} sin(3t)/t
Change variables! Let x = 3t. Then t = x/3 and we have

lim_{\t \to 0} sin(3t)/t
= lim_{x \to 0} sin(x)/[x/3]
= lim_{x \to 0} 3 * sin(x) / x
= 3 * lim_{x \to 0} sin(x) / x
= 3 * 1
= 3

Substituting back in to our above we have again that
lim_{t \to 0} sin^2(3t)/(t^2)
= 3^2 = 9

Answer 4

u need to specify if its (sin^2 * 3t)/ t^2 or (sin^2) * (3t/t^2) I think u were aiming for the second one, but the way you wrote it order of operations will change the answer.

Answer 5

Try graphing it on a TI and see what happens as x approaches zero. IT's a good way to learn more about limits.

Answer 6

Go to bittorent and download a calculas book! Sorry i could'nt help you im still in elemtary algebra but my husband has taken and passed all 3 calculas classes .