# How would I find the limit as t -> 0 of tan(6t)/sin(2t)?

Here it is again: find the limit as t approaches 0 of

tan(6t)/sin(2t)

I'm guessing it has something to do with the identity lim x->0 of sin(x)/x = 1, but I don't know how to use it in this problem, since the numbers within tan and sin are different.

Yes this can all be done with lim x->0 of sin(x)/x = 1. Start by dividing top and bottom by 2t:

lim(t-->0) [tan(6t) / 2t] / [sin(2t) / 2t]

Now the denominator goes to 1:

lim(t-->0) [tan(6t) / 2t] / 1

lim(t-->0) tan(6t) / 2t

Remember that tan(6t) = sin(6t) / cos(6t):

lim(t-->0) [sin(6t) / cos(6t)] / 2t

lim(t-->0) sin(6t) / [2t ? cos(6t)]

Divide top and bottom by 6t:

lim(t-->0) [sin(6t) / 6t] / [(2t ?cos(6t)) / 6t]

You get 1 in the numerator:

lim(t-->0) 1 / [(2t ?cos(6t)) / 6t]

lim(t-->0) 6t / [2t ?cos(6t)]

lim(t-->0) 6 / 2cos(6t)

lim(t-->0) 3 / cos(6t)

Now you can plug in 0:

3 / cos(0)

= 3 / 1

Oct 9 at 5:44

limit as t -> 0 of tan(6t)/sin(2t) = 6/2 = 3

--------

Ideas: tan(6t) ~6t, and sin(2t) ~ 2t as t -> 0

Oct 9 at 9:30

d(tan6t)/dt = 6sec26t = 6/cos26t

d(sin2t)/dt = 2cos2t

Using L'Hopital's Rule:

lim t->0 tan(6t)/sin(2t) = lim t->0 3/cos26tcos2t = 3

...

Oct 9 at 13:39