anonymous
  • anonymous
find the limit
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • anonymous
taught*
ksaimouli
  • ksaimouli
1

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I need to show algebraically
anonymous
  • anonymous
we were taught to set it up as an inequality so \[-1\le \frac{ \sin2x }{ x }\le1 \]
anonymous
  • 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
  • anonymous
and wait it is 2 not 1
anonymous
  • anonymous
so replace the x^2+1 with x^2+2 and same with the negative
anonymous
  • anonymous
sorry it should be \[-x ^{2}-2\]
anonymous
  • anonymous
Ok I get the first thing but the next two what?
Psymon
  • 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
  • anonymous
multiply throughout by 2 but doesnt that just leve me with sin?
Psymon
  • Psymon
|dw:1378861537916:dw|
anonymous
  • anonymous
wait I dont under stand what you are canceling because wouldn't that give you |dw:1378861614653:dw|
Psymon
  • 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
  • anonymous
wait is that a trig identity? how do you know that equals 1?
Psymon
  • 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
  • anonymous
I do not remember learning that but I trust you
anonymous
  • 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
  • Psymon
Well, this is the squeeze theorem.
Psymon
  • 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
  • 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
  • Psymon
But you learned the whole 1 <= sinx/x <= 1 kinda thing, right?
anonymous
  • anonymous
yes
Psymon
  • 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
  • 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
  • anonymous
so I would write that as \[\lim_{x \rightarrow 0}2*1=2\]
Psymon
  • Psymon
Yep, absolutely. Just what you said on both your posts. So yes, 2*1 = 2
anonymous
  • anonymous
alright thank you are you ok if I ask you about one more same type of problem|?
Psymon
  • Psymon
Go for it.
anonymous
  • anonymous
\[\frac{ \frac{ 1 }{ 2+x }-\frac{ 1 }{ 2 } }{ x }\]
anonymous
  • anonymous
same limit as x approaches 0
anonymous
  • anonymous
actually I need help on a number more
anonymous
  • anonymous
all the same type
Psymon
  • 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
  • anonymous
ok that one I see alright the next is \[\frac{ 3\sin 4x }{ \sin 3x }\]
Psymon
  • 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
  • anonymous
well could it possibly become\[\frac{ 3*4sinxcosx }{ 2sinxcosxsinx }\] we cross out and are left with\[\frac{ 6 }{ sinx }\]
Psymon
  • Psymon
Lemme check the identity work you did xD
Psymon
  • Psymon
Bleh, stupid identities, haha.
Psymon
  • Psymon
Alright, I got it x_x
anonymous
  • anonymous
how did you get x_x and what operation is _
anonymous
  • anonymous
jk
Psymon
  • 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
  • Psymon
|dw:1378865214777:dw| |dw:1378865597674:dw| Now just plug in 0 finally x_x
anonymous
  • anonymous
alright thanks now last one \[\frac{ x+sinx }{ x }\] same as x approaches 0
Psymon
  • Psymon
You can split it into two fractions: |dw:1378866255817:dw| Pretty straightforward from there.
anonymous
  • anonymous
but the limit is zero so I would get xsinx+1 which is one
anonymous
  • anonymous
sorry the limit is 2 I mean
anonymous
  • anonymous
so I have the lim=2 but 1+xsinx yields one
Psymon
  • Psymon
Well, x/x = 1 in the first fraction and the sandwich identity that says sinx/x is 1. SO 1 + 1 = 2.
anonymous
  • 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
  • Psymon
Yeah, np. Night :3

Looking for something else?

Not the answer you are looking for? Search for more explanations.