What is the limit of tan 6t/sin 2t as t approaches 0?

Question

I'm confused...should I use the Quotient Rule or conjugation?HELP?!!!!

Answer 1

lim (tan6t / sin2t)
t -> 0

One thing you can do is use L'Hospital's rule. Since you get the form [0/0], you can do this. Take the derivative of the top and the bottom for a new limit.

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

Next, pull out the 6 on top and the 2 on the bottom, outside of the limit. This pulls out as 6/2, or 3.

3 * lim ( sec^2(6t) / cos(2t) )
t -> 0

Now, plug in the value of 0 directly.

3 * sec^2(0) / cos(0)
3 * (1)^1 / 1
3 * (1)
3

Your answer should be 3.

Answer 2

in case you in basic terms plug interior the shrink you get the indeterminate type 0/0, so l'hopital's rule may well be utilized 6sec²(6t) / 2cos(2t) Now if we plug it in we get 6/2 = 3, because of the fact sec(0) = cos(0) = a million

Answer 3

Use L'Hopitals Rule (i.e. differentiate the top and bottom of the fraction and see what happens)

d/dt(tan 6t/sin 2t)
=(6sec^2 6t/2cos 2t)

we can plug 0 into this now
if we do, we get:

6/2

=3

Answer 4

Here is a simple way,

lim6t/sin 2t
t→0

= lim6t/2t, sin2t~2t
.. t→0

= 3
--------
Reason: sin2t has the same order to approach zero as 2t.

Answer 5

The limit is 3 because = lim3* 2t/sin2t