find the limit

- anonymous

find the limit

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- anonymous

\[\lim_{x \rightarrow 0}\frac{ \sin 2x }{ x }\]
We were taight the sandwhich theorem but I am not sure how to do it for this limit

- anonymous

taught*

- ksaimouli

1

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## More answers

- anonymous

I need to show algebraically

- anonymous

we were taught to set it up as an inequality so
\[-1\le \frac{ \sin2x }{ x }\le1 \]

- anonymous

and then multiply through by the equation you sandwich it in and it would work for \[x ^{2}+1\]\[x ^{2}-1\] but cant you multiply through by only one equation?

- anonymous

and wait it is 2 not 1

- anonymous

so replace the x^2+1 with x^2+2 and same with the negative

- anonymous

sorry it should be \[-x ^{2}-2\]

- anonymous

Ok I get the first thing but the next two what?

- Psymon

All you need is the angle of sin and something in the denominator to match. If you have sin(2x), then you need 2x in the denominator somehow. So how do you get a 2 in the denominator?

- anonymous

multiply throughout by 2 but doesnt that just leve me with sin?

- Psymon

|dw:1378861537916:dw|

- anonymous

wait I dont under stand what you are canceling because wouldn't that give you |dw:1378861614653:dw|

- Psymon

Yeah, exactly. But all we need is the angle of sin to match something in the denominator. So if we have 2x in sin, we need 2x in the denominator. So yes, thats exactly what youd have. The identity is:
\[\lim_{x \rightarrow 0}\frac{ sinx }{ x } =1 \]Well, we now have just that. We have 2x in sin and 2x on bottom, so:
|dw:1378861718546:dw|

- anonymous

wait is that a trig identity? how do you know that equals 1?

- Psymon

No, its a limit identity. Its just in your textbook or when youre initially shown it, youre only shown it as sinx/x. But as long as those two x's match it works the same way. Hell, I can have this:
\[\lim_{x \rightarrow 0}\frac{ \sin(\sin ^{\pi*e}(x)) }{ \sin ^{\pi *e}(x) }=1 \]
Even though it looks ridiculous,it fits thepattern. The angle of sin in the numerator matches whats in the denominator, so the whole thing is one.

- anonymous

I do not remember learning that but I trust you

- anonymous

but I feel like there is another way because what we were taught was the "sandwhich theorm" so i feel like there is some equation like x^2 I need to multiply by and that will help me solve it

- Psymon

Well, this is the squeeze theorem.

- Psymon

Its just we dont HAVE to put inequalities and do somesort of proof. We just have to match the limit to what we know the identity is.

- anonymous

he said that the squeeze is just another name and my problem is that I dont recognize or remember learning that identity so I want to understand it not just know that sin2x/2x=1 because I might not remember that

- Psymon

But you learned the whole 1 <= sinx/x <= 1 kinda thing, right?

- anonymous

yes

- Psymon

Right, its the same thing, as long as those x's up there match
1 <= sin3x/3x <=1
1<= sin675x/675x <= 1
Theyre all the same thing.

- anonymous

so its that if we have the variable multiplied by the same number on top inside the sine as the variable multiplied on bottom it equals one?

- anonymous

so I would write that as
\[\lim_{x \rightarrow 0}2*1=2\]

- Psymon

Yep, absolutely. Just what you said on both your posts. So yes, 2*1 = 2

- anonymous

alright thank you are you ok if I ask you about one more same type of problem|?

- Psymon

Go for it.

- anonymous

\[\frac{ \frac{ 1 }{ 2+x }-\frac{ 1 }{ 2 } }{ x }\]

- anonymous

same limit as x approaches 0

- anonymous

actually I need help on a number more

- anonymous

all the same type

- Psymon

Alright, here's how you can maybe work it out. Realize this:
|dw:1378863398301:dw|
So try multiplying it through and see if you can simplify :3

- anonymous

ok that one I see alright the next is \[\frac{ 3\sin 4x }{ \sin 3x }\]

- Psymon

Ive seen thisproblem before, lol. Maybe coincidence. Okay, so as you can see the angle in top is 4x. Problem is theres no way youre going to get a 4x in the bottom, that sin(3x) is just stuck there. This means we're forced to use identities to see if we can simplify it.

- anonymous

well could it possibly become\[\frac{ 3*4sinxcosx }{ 2sinxcosxsinx }\] we cross out and are left with\[\frac{ 6 }{ sinx }\]

- Psymon

Lemme check the identity work you did xD

- Psymon

Bleh, stupid identities, haha.

- Psymon

Alright, I got it x_x

- anonymous

how did you get x_x and what operation is _

- anonymous

jk

- Psymon

Oh, its just an emote xD And yeah, tons of identities, unfortunately. The idea is you break it down until you get to a point where its safe to plug in 0 finally. So Ill do a few of them.

- Psymon

|dw:1378865214777:dw|
|dw:1378865597674:dw|
Now just plug in 0 finally x_x

- anonymous

alright thanks now last one \[\frac{ x+sinx }{ x }\] same as x approaches 0

- Psymon

You can split it into two fractions:
|dw:1378866255817:dw|
Pretty straightforward from there.

- anonymous

but the limit is zero so I would get xsinx+1 which is one

- anonymous

sorry the limit is 2 I mean

- anonymous

so I have the lim=2 but 1+xsinx yields one

- Psymon

Well, x/x = 1 in the first fraction and the sandwich identity that says sinx/x is 1. SO 1 + 1 = 2.

- anonymous

alright thank you so much im going to make sure to have my teacher go over all this too and you helped so much I really appreciate all your help goodnight

- Psymon

Yeah, np. Night :3

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