Limit problems - solving without L`Hopital's rule?

Question

1) lim t-->0 tan(6t) / sin(3t)
2) lim x-->0 (sin^2(4t)) / t^2

I posted this the other day but forgot to specify that we haven't learned L'Hopital's rule yet. So how would I solve these problems otherwise? I'm stuck.

Answer 1

Theorem:
lim sint / t = 1 (as t → 0)

this is usually the theorem that is used in more basic calculus, although the proof is slightly complicated.

using that:

lim tan[6t]/sin[3t] = lim tan[6t]/t / sin[3t]/t
= lim {sin[6t]/t} / c0s[6t]* {sin[3t]/t}
= lim {sin[6t]/t} * lim 1/ cos[6t] * 1/lim {sin[3t]/t}
= lim 6*{sin[6t]/ 6t} * lim 1/ cos[6t] * 1/lim 3{sin[3t]/3t}
= 6*1*1/3 = 2

lim [sin(4t)/4t]^2 * 16 = 16


trick: whenever you got a
lim sin(px)/x ... you multiply and divide by p to make it:
p lim sin(px)/px .... then lim sin(px) / px =1...
the result of lim sin(px)/x = p.


Also special note: this is usually included in this discussion:
lim (1-cos[t]) / t = 0 ... as t→0.

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Answer 2

L Hopital Calculator

Answer 3

The trick to limits of trig functions is to express them more in terms of cos x, because it is 1. So a little trig manipulation for the first one will give us the answer:

let 3t = x, 6t=2x and the problem becomes:

tan(2x) / sin x

Now, from the double angle formula we have:
cos 2x = cos² x - sin² x
sin 2x = 2 sin x cos x

so we get:
tan 2x / sin x = [sin 2x / cos 2x] / sin x
(2 sin x cos x) /(cos² x - sin² x) / sin x
2 cos x / (cos² x - sin² x)
as t approaches 0 so does x, and so cos x approaches 1 and sin x approaches 0 and we wind up with:

2 (1) /(1-0) = 2

for the second one:

(sin² 4t) / (t²) = [ (sin 4t) / t ]² = [ (sin 2(2t)) / t ]²

using the double angle formula for angle 2t we get:

[ (sin 2(2t) ) / t ]²
[ (2 sin 2t cos 2t) / t ]²

using double angle again we get:
[ (2 (2 sin t cos t) (cos² t - sin² t)) / t ]²
[ (4)(sin t /t) (cos t) (cos² t - sin² t)]²

Now, using the well known limit [(sin t ) / t ] = 1 as t → 0.
(see the proof in your calc book) we get:

[4(1) (1) (1-0)]² =[(4)]² = 16

Answer 4

As x --> 0, tan x --> x
Also, as x--> 0, sin x --> x and cos x --> 1

So, the 1st problem becomes 6t/3t = 2

2nd problem, (4t)^2/t^2 = 16

Answer 5

The first one is easy; tan(x) = sin(x)/cos(x). You have the added complication of needing to use the double angle formula for tangent (look it up). [Simplify your life by letting z = 3t; it won't affect the limit.] The second one isn't obvious. I ran into a problem like this in a freshman calculus final, when l'Hopital's rule hadn't yet been introduced; I answered the question by proving l'Hopital on the spot (it's trivial), and then using it. (Got an A.)

Answer 6

1) lim t-->0 tan(6t) / sin(3t) = 6t/(3t) = 2
2) lim t-->0 (sin^2(4t)) / t^2 = (4t)^2/(t^2) = 16
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Ideas: As x-->0, tan(x)~sin(x)~x.