Question

find the limit as x approaches 0 of x/sin3x....i need somebody to explain and show steps...thanks :)

Answer 1

we first realize or utilize the fact that:
sin t
lim ----- = 1
t->0 t

Answer 2

i take it that this is your problem:

x
lim --------
x->0 sin 3x

Answer 3

yes that's right

Answer 4

okay,
now here is what we will do, we substitute t = 3x.

Answer 5

ok

Answer 6

then we notice that if t = 3x then x = t/3.

Answer 7

ok, i get that part

Answer 8

Now, if x -> 0, then 3x -> 0, so t -> 0

Answer 9

which then gives us:

t/3
lim -------
t->0 sin t

Answer 10

alright

Answer 11

algebra will take care of the rest

Answer 12

I'm just confused as to how to find the answer when the sin is involved

Answer 13

okay, hold on we are not done yet

Answer 14

i understand that you multiply the numerator by the reciprocal of the denominator and i got 1/3sin

Answer 15

now, :
t/3
lim -------
t->0 sin t

becomes to:


t
lim ------- =
t->0 3 sin t

Answer 16

then:


1 t
lim - ------ =
t->0 3 sin t

Answer 17

jeffries, this is the same thing I showed you before, use L'Hopital's Rule

Answer 18

no, L'Hospital Rule needed here

Answer 19

ooh ok

Answer 20

continuing:
1 1
lim - -------- =
t->0 3 sin t/t

Answer 21

if you have x/sin(3x), when x approaches 0, you get the 0/0 indeterminate form again

Answer 22

1 1
- lim -------- =
3 t->0 sin t/t

Answer 23

1 1
- ------------------ =
3 lim [t->0] sin t/t

Answer 24

1 1 1
- --- = ---
3 1 3

Answer 25

so you take the derivative of top and bottom, with respect to x, this gives:
1/3cosx, now you sub in x =0 again:
1/3*cos(0), since cos(0) = 1, your answer is 1/3

Answer 26

so the answer is 1/3?

Answer 27

yes answer is 1/3

Answer 28

thanks so much

Answer 29

you dont need L'Hospital rule here. Obviously camp2001 has a one track mind