Lim as t--->0 (tan 6t)/(sin 2t)
@Luigi0210

- anonymous

Lim as t--->0 (tan 6t)/(sin 2t)
@Luigi0210

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- Luigi0210

Abb0t, get this?

- anonymous

huh?:o

- abb0t

Have you learned L'Hopitals rule?

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## More answers

- anonymous

I don't know can you explain it to me please? do you learn it in calc 1?

- anonymous

You do eventually. Do you know about derivatives?

- abb0t

Try changing tangent to sine and cosines.

- anonymous

No not yet :/ I heard it is much simpler but we haven't learned it yet !

- abb0t

Then, sin(2t) to 2sin(t)cos(t)

- abb0t

You should see then.

- anonymous

but what would happen to the 6t?? when I make the tan sin and cos?

- abb0t

sin(6t) = 2sin(t)cos(t)(2cos(2t)-1)(2cos(2t)+1)

- 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

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

& the problem was different because it didn't involve tangent only sin so of course it was simpler

- anonymous

I am better at finding limits when its not involving trig :(

- 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

There's got to be a rule for \(\displaystyle \lim_{t\to0}\frac{\sin at}{\sin bt}\)...

- anonymous

how did you get \[=1/\cos6t\]

- anonymous

like what happened to sin2t

- 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

The \(\sin 2t\) is still there.

- anonymous

the rules my teacher gave us were lim \[\theta-->0 \sin \theta \div \theta \] = 1

- anonymous

ohhh ok I see what you did! are you trying to cancel out the sin??

- 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

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

ohhh ok i see what you did that is tedious!! lol but thank you sooo much!!:)

- anonymous

You're welcome!

- anonymous

so basically for this question you have to split everything up and multiply to cancel:o

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