Question

Hi everyone! Please explain how to find the limit as "t" goes to infinity of (-sin(2t)) / (se^(st))...this looks like an improper/improper which needs L'ophital's rule but not sure how to implement that. Anyone? Thanks! :o)

Answer 1

now that I think about it, if sin(2t) bounces back and forth forever between y=-1 and y=1, then the numerator is a real number and the denominator goes to infinity, which would make the lim go to zero correct?

Answer 2

thats right only if s > 0

Answer 3

HI!!

Answer 4

well how would I know if s>0?

Answer 5

you very much seem to be doing laplace transforms ?

Answer 6

lol...yes! :o)

Answer 7

then s > 0 is assumed

Answer 8

this may seem like a dumb question but, why is it assumed?

Answer 9

you wont ask that question if you look at the integral that you use to find laplace transform

Answer 10

not sure I understand what you mean

Answer 11

look at the definition of laplace transform in your notes

Answer 12

and ask yourself for what values of "s" the integral makes sense (converges)

Answer 13

oh yes....I remember...integral is between zero and infinity, so it's assumed that s>0

Answer 14

actually, that wasn't a dumb question, it helped me remember the root of what I'm doing!

Answer 15

it is a good question, i didnt answer directly because i want you look at that laplace transform integral haha

Answer 16

if s is a negative, then the LaPlace definition diverges! :O)

Answer 17

as simple as that

Answer 18

yay!...now back to my original problem...this was only a portion of one of those nasty ones where you have to set your work equal to the original laplace transform and do all the tricky cancelling! bleh! :o/

Answer 19

Thanks rational!

Answer 20

you're doing good, yes that limit evaluates to 0

Answer 21

yw and good luck!

Answer 22

If you need to justify well, you may use squeeze thm

Answer 23

\[-1\le\sin(2t)\le1\]

Answer 24

i don't even remember squeeze theorem...but not sure I can take the time to relearn right now...maybe later

Answer 25

kk squeeze thm is for real analysis, rigor is not needed when you're in diff eqns ;p

Answer 26

good! whew!

Answer 27

I will be doing math for the next 7 hours...will you be around?

Answer 28

Yes feel free to tag me :)

Answer 29

thanks @rational

Answer 30

haha

Answer 31

quick question however...
do you know what those weird squiggly "u's" are for?

Answer 32

unit step functions ?

Answer 33

something about step functions maybe?

Answer 34

yes I think so

Answer 35

can you explain those at all...like what is the purpose of the symbol?

Answer 36

it is a piecewise function :
\[\large u_c(t)=\left\{\begin{array}{}0&:&t\lt c\\1&:&t\ge c\end{array}\right.\]

Answer 37

for example, \(\large u_0(t)\) looks like this :