Question

Limit for medal :D ?

Answer 1

\[\lim_{t \rightarrow 0^{+}} \frac{ 3\sin(t)-3\sin(3t) }{ 3\tan(t)-3\tan(3t) }\]

Answer 2

Hmmm, first of all you can cancel out all the threes.

Answer 3

And then, I assume you do l'hopital's rule maybe?

Answer 4

nope :P

Answer 5

Well... maybe you could!

Answer 6

But yeah :P L'Hopital's rule

Answer 7

write in terms of sine and cosine
also divide top and bottom by 3x

Answer 8

use \[\lim_{u \rightarrow 0} \frac{\sin(u)}{u}=1 \]

Answer 9

which means
\[\lim_{x \rightarrow 0}\frac{\sin(3x)}{3x}=1\]

Answer 10

nope... it's a bunch of horrible work *Hint* XD @myininaya

Answer 11

have you tried it....

Answer 12

do you not know how to write in terms of sine and cosine?

Answer 13

tan(x)=sin(x)/cos(x)

Answer 14

Yeah I have the answer/work

Answer 15

ok did you want me to check your work?

Answer 16

@myininaya ...
\[\lim_{t \rightarrow 0+}\frac{ 3\sin(t)-3\sin(3t) }{ 3\tan(t)-\tan(3t) }\]
*Let
\[f(t)=3\sin(t)-\sin(3t)\]
*Let
\[g(t)=3\tan(t)-\tan(3t)\]

f(0)=0
g(0)=0
*Can use l'hopital's rule

\[f'(t)=3\cos(t)-\cos(3t)\]
\[g'(t)=3\sec^2(t)-3\sec^2(3t)\]
f'(0)=0
g'(0)=0
*Can use l'hopital's rule again


\[f''(t)=-3\sin(t)+9\sin(3t)\]
\[g''(t)=6\sec^2(t)tan(t)-18\sec^2(3t)tan(3t)\]
f''(0)=0
g''(0)=0
*Use l'hopital's rule again

\[f'''(t)=-3\cos(t)+27\cos(3t)\]

\[g'''(t)=-54sec^4(3t)-108sec^237(tan^2(3t)+6sec^4(t)+12sec^2(t)tan^2(t)\]
f'''(0)=24
g'''(0)=-48

Answer 17

oh mine

Answer 18

i will show you the way i was thinking
it doesn't include l'hospital
but if it must include l'hospital i can go through and see if i can find an error

Answer 19

Than...

\[\lim_{t \rightarrow 0^+}\frac{ 3\sin(t)-\sin(3t) }{ 3\tan(t)-\tan(3t) }=\lim_{t \rightarrow 0^+}\frac{ f'(t) }{ g'(t) }=\lim_{t \rightarrow 0^+}\frac{ f''(t) }{ g''(t) }=\lim_{t \rightarrow 0^+}\frac{ f'''(t) }{ g'''(t) }=\frac{ 24 }{ -48 }=\frac{ -1 }{2 }\]

:D
And okay sounds good @myininaya
I'd like to see how you do that

Answer 20

** Corrections:
\[f'(t)=3\cos(t)-3\cos(3t)\]

Answer 21

\[\lim_{t \rightarrow 0^{+} } \frac{ 3\sin(t)-3\sin(3t) }{ 3\tan(t)-3\tan(3t) } \\ \lim_{t \rightarrow 0^+} \frac{3 \sin(t)-3 \sin(3t)}{3 \frac{\sin(t)}{\cos(t)}-3 \frac{\sin(3t)}{\cos(3t)}} \text{ now divide both \top and bottom by } 3t \\ \\ \lim_{t \rightarrow 0^{+}} \frac{ 3\frac{\sin(t)}{\color{red}{3t}}-3\frac{\sin(3t)}{\color{red}{3t}} }{ 3\frac{\sin(t)}{\color{red}{3t}\cos(t)}-3\frac{\sin(3t) }{\color{red}{3t}\cos(3t)}} \\ \lim_{t \rightarrow 0^+} \frac{\frac{\sin(t)}{t}-3 \frac{\sin(3t)}{3t}}{\frac{\sin(t)}{t} \frac{1}{\cos(t)}-3 \frac{\sin(3t)}{3t} \frac{1}{\cos(3t)}} \\ =\frac{1-3(1)}{1\cdot \frac{1}{1}-3 \cdot 1 \cdot \frac{1}{1}} \\ =\frac{1-3}{1-3} \\ \frac{-2}{-2} \\ =1\]

Answer 22

in your orinigal problem you had a 3 next to the tan(3t) in bottom

Answer 23

i think in your second one you wrote you had no 3 next to the tan(3t) in bottom

Answer 24

oh i'm sorry lol

Answer 25

so no 3

Answer 26

No oops XD your right * I was thinking of the above ^^ ah

Answer 27

your limit is different from mine

Answer 28

i will look at your work now

Answer 29

Okay thank you (: @myininaya

Answer 30

if f(t)=3sin(t)-sin(3t)
then
f'(t)=3 cos(t)-3 cos(3t)
you have f'(t)=3 cos(t)-cos(3t)

Answer 31

f'(0)=3(1)-3(1) is 0
so i think that is what you meant anyways

Answer 32

I put the correction up there ^
D: sorry i always forget numbers haha

Answer 33

ok you answer looks good
\[\lim_{t \rightarrow 0^+} \frac{3 \sin(t)-\sin(3t)}{3 \tan(t)-\tan(3t)}= \frac{-1}{2} \\\]

Answer 34

Cool, thanks for showing me that ^ :D

Answer 35

well after changing the problem to that I don't think my way will work since it will give
(1-1)/(1-1) which is indeterminate form

Answer 36

Hmm.. yeah didn't try it that way/think about trying it that way soo not sure!
It's pretty easy... just do L'hopital's rule 3 x

Answer 37

This is not by
definition @sunnnystrong are you sure you wan't \(\epsilon-\delta\) proofs?

Answer 38

@zzr0ck3r yeah just answer any way you want? XD

Answer 39

It will take me a while to figure them out and I don't want to unless it is worth it. Do you know what I mean by epsilon delta proofs? Do you know the epsilon- delta definition of a limit? It is quite involved and not so easy with some trig functions.

Answer 40

Wellll @zzr0ck3r i am asking only for you to evaluate the limit & find the numerical value.

However, i'd love to learn more about the epsilon-delta definition of a limit. :O (which yes, looks time consuming but i will definitely listen!)

Answer 41

It's the first 3 weeks of a analysis class, it's to much to teach here. But google it and you will find some good stuff and it can be quite eye opening if you understand the point.