Limit.......?

Would the answer to this problem be 0?





lim as t--%26gt;0 (sin^2 3t)/(t^2)|||No, the answer would be 9.





As you know, substituting the value t = 0 directly into this will not work so an alternate method is needed.





There are some special trigonometric limits you would need to know to solve this:





1) Lim as (x --%26gt; 0) {(sin x)/x} = 1


2) Lim as (x --%26gt; 0) {(1 - cos x)/x} = 0





You only need the first one to solve this one so here goes:





First you want to write the expression in the same form as the special trig limit #1 above. So you have





Lim as t--%26gt;0 (sin^2 3t)/(t^2) = Lim as t--%26gt;0 [{(sin 3t)/t} * {(sin 3t)/t}]





This is almost like that form but the denominator has to be the same as the quantity you are looking for the sin of, so it becomes





Lim as t--%26gt;0 9[{(sin 3t)/3t} * {(sin 3t)/3t}]





Note that 9[{(sin 3t)/3t} * {(sin 3t)/3t}] is an equivalent expression to


[{(sin 3t)/t} * {(sin 3t)/t}]





You can now go ahead and use the special trig limit to get





Lim as t--%26gt;0 9[{(sin 3t)/3t} * {(sin 3t)/3t}] = 9 * Lim as t--%26gt;0 [{(sin 3t)/3t} * {(sin 3t)/3t}]





= 9 * 1 * 1 = 9


since we know from the properties of limits that,


Lim as x --%26gt; c {b f(x)} = bL, where b and c are real numbers and L is the limit of f(x) as x--%26gt; c.





Therefore, lim as t--%26gt;0 (sin^2 3t)/(t^2) = 9|||no, 0 would be an asymptote|||No. You'll end up with sin(0)/0, which is 1/0. It's undefined.|||It's not undefined. Take l'hopitals. I get 9. There's also this property where lim as x--%26gt;0 sinx/x = 1 so im as x--%26gt;0 sin3x/x = 3. In this problem it's squared so 3 squared = 9.|||the answer is not undefined gosh its simple.





lim (sin^2 (3t))/t^2 = lim (sin(3t)/t ) * lim (sin(3t)/t)


t-%26gt;0 t-%26gt;0 t-%26gt;0





= 3*3


=9