quick question... how do i find the limit of the function tan 6t/sin 2t as t approaches 0...?

Question

i did

sin 6t/cos6t/sin 2t...


umm.. if im going in the right track... should it be (sin 6t)(sin 2t)/cos6t... or should it be...
sin6t/(cos 6t)(sin 2t)???

Answer 1

It's sin(6t)/[cos(6t) sin(2t)], and as such the limit is

6t/2t = 3.

Purists will tell you to express the original function as:

6t [sin(6t)/(6t)] / [cos(6t) 2t {sin(2t)} / (2t)], so that for both sine terms and their associated divisors you can appeal to the standard result that as x --> 0, sin x / x --> 1. Then, since cos (6t) --> 1, one has left just the additional factors that were introduced, namely

6t / 2t = 3.

But all of that is of course just pedantic mathematical icing on the cake. Since even the result that sin x / x --> 1 is based upon knowing the EXPANSION of sin x (= x - x^3 / 3! + ...), one might as well appeal to the leading term of sin nt being nt and have done with it!

Live long and prosper.

Answer 2

Since the expression become 0/0 at the limit, use L'hospital's rule, and take the derivative of the numerator and denominator separately

d/dx (tanx) = sec^2(x)*dx/dt
d/dx (siny) =cos(y)*dy/dt

Set x = 6t, y = 2t:

d/dt (tan6t) = 6*sec^2(6t)
d/dt (sin2t) = 2*cos2t

sec^2 = 1/cos^2, so sec^2(6t) goes to 1 at t = 0
cos(2t) at t=0 is also 1

the limit is then 6/2 or 3

Answer 3

Use L'Hopitals Rule (i.e. differentiate the right and backside of the fraction and see what happens) d/dt(tan 6t/sin 2t) =(6sec^2 6t/2cos 2t) we are able to plug 0 into this now if we do, we get: 6/2 =3

Answer 4

it's 3.

lim of (tan 6x / 6x) = 1
lim of (2x / sin 2x) = 1
using the lim of [ (tan 6x / 6x) * (2x / sin 2x) * (6x / 2x)] = 3