I need help with a math question involving finding a limit

Q: Find the limit lim as t -> 0 is [sin(3t)*sin(5t)] / t^2

Can someone explain how to find this limit?

you know the common limit concept:
lim(x->0) sin(x) / x = 1 right?

you try and construct that same idea.


lim(t->0) [sin(3t)sin(5t)] / t^2
Try and think what would need to be on the denominator to "cancel" the sine's out (not litearlly cancel out but conceputally, approach 1).
So, to "cancel" the sin(3t) you need a 3t on the dnominator, fort he sin(5t) you need a 5t on the denominator...since you can't just input 15t^2 on the denomintaor, you will have to multiply top and bottom by 15t^2.

= lim [sin(3t)sin(5t)*15t^2] / [(3t)(5t)t^2]
= lim [sin(3t)/(3t)][sin(5t)/(5t)][15t^2 / t^2]

sin(3t)/3t --> 1
sin(5t)/5t --> 1
t^2's literally cancels out. and we're left with...

= (1)(1)(15)
= 15

Hope this helps :D

Use the special limit theorem : lim(x->0) sinx/x= 1

Lim(t->0)[sin(3t)*sin(5t)]/t^2=

Lim(t->0)[ sin(3t)/(3t)] * [sin(5t)/(5t)] * (3*5)

= (1)(1)((15)

=15

Hoping this helps!