anonymous
  • anonymous
Lim as t--->0 (tan 6t)/(sin 2t) @Luigi0210
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Luigi0210
  • Luigi0210
Abb0t, get this?
anonymous
  • anonymous
huh?:o
abb0t
  • abb0t
Have you learned L'Hopitals rule?

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anonymous
  • anonymous
I don't know can you explain it to me please? do you learn it in calc 1?
anonymous
  • anonymous
You do eventually. Do you know about derivatives?
abb0t
  • abb0t
Try changing tangent to sine and cosines.
anonymous
  • anonymous
No not yet :/ I heard it is much simpler but we haven't learned it yet !
abb0t
  • abb0t
Then, sin(2t) to 2sin(t)cos(t)
abb0t
  • abb0t
You should see then.
anonymous
  • anonymous
but what would happen to the 6t?? when I make the tan sin and cos?
abb0t
  • abb0t
sin(6t) = 2sin(t)cos(t)(2cos(2t)-1)(2cos(2t)+1)
abb0t
  • abb0t
this problem is just long and tedious w/o l'hopitals rule lol but there are certain rules. also, where you know that I think some limit of sine = 1 and some limit of cosine = 0 or 1?
anonymous
  • anonymous
lol I know that's why i'm getting really lost because I don't recall L'hopitals rule and I don't know the steps of how to work it out.. :( my teacher gave an example of a similar problem but I don't completely understand it
anonymous
  • anonymous
& the problem was different because it didn't involve tangent only sin so of course it was simpler
anonymous
  • anonymous
I am better at finding limits when its not involving trig :(
anonymous
  • anonymous
\[\frac{\tan6t}{\sin2t}=\frac{\frac{\sin6t}{\cos6t}}{\sin2t}=\frac{1}{\cos6t}\cdot\frac{\sin6t}{\sin2t}\] Tedious indeed... Double angle identity will help a bit.
anonymous
  • anonymous
There's got to be a rule for \(\displaystyle \lim_{t\to0}\frac{\sin at}{\sin bt}\)...
anonymous
  • anonymous
how did you get \[=1/\cos6t\]
anonymous
  • anonymous
like what happened to sin2t
anonymous
  • anonymous
It came from the tangent. \(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles \[\frac{\tan6t}{\sin2t}=\frac{\color{red}{\frac{\sin6t}{\cos6t}}}{\sin2t}=\frac{1}{\cos6t}\cdot\frac{\sin6t}{\sin2t}\] \(\color{blue}{\text{End of Quote}}\) Then some algebraic rearrangement to make it look nicer.
anonymous
  • anonymous
The \(\sin 2t\) is still there.
anonymous
  • anonymous
the rules my teacher gave us were lim \[\theta-->0 \sin \theta \div \theta \] = 1
anonymous
  • anonymous
ohhh ok I see what you did! are you trying to cancel out the sin??
anonymous
  • anonymous
Oh, I see. Much simpler than what I was thinking. So you know that \[\lim_{t\to0}\frac{\sin t}{t}=1\] and that, more generally, \[\lim_{t\to0}\frac{\sin at}{at}=1\] Now, having rewritten the original limit a bit, let's deal with what we have: \[\lim_{t\to0}\frac{1}{\cos6t}\cdot\frac{\sin6t}{1}\cdot\frac{1}{\sin2t}\] The limit of a product is the same as the product of the limits; mathematically, you have \[\lim_{t\to0}\frac{1}{\cos6t}\cdot\lim_{t\to0}\frac{\sin6t}{1}\cdot\lim_{t\to0}\frac{1}{\sin2t}\] The first limit is pretty straightforward. For the second limit, you do the following: \[\lim_{t\to0}\frac{\sin6t}{1}\cdot\frac{6t}{6t}=\lim_{t\to0}\color{red}{\frac{\sin6t}{6t}}6t=\lim_{t\to0}\frac{\sin6t}{6t}\cdot\lim_{t\to0}6t=\lim_{t\to0}6t\] And for the third limit, you do something similar: \[\lim_{t\to0}\frac{1}{\sin2t}\cdot\frac{2t}{2t}=\lim_{t\to0}\color{blue}{\frac{2t}{\sin2t}}\cdot\frac{1}{2t}=\lim_{t\to0}\frac{2t}{\sin2t}\cdot\lim_{t\to0}\frac{1}{2t}=\lim_{t\to0}\frac{1}{2t}\]
anonymous
  • anonymous
So, you have \[\lim_{t\to0}\frac{1}{\cos6t}\cdot\lim_{t\to0}6t\cdot\lim_{t\to0}\frac{1}{2t}=\color{red}{\lim_{t\to0}\frac{6t}{2t}}\]
anonymous
  • anonymous
ohhh ok i see what you did that is tedious!! lol but thank you sooo much!!:)
anonymous
  • anonymous
You're welcome!
anonymous
  • anonymous
so basically for this question you have to split everything up and multiply to cancel:o

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