I had a quiz today, apparently I got this question wrong, but I'm pretty sure I'm right. I was going to ask my teacher about it tomorrow in class, but I want to make sure my answer is reasonable first.

What's the limit of f(x) = (sin(3x))/2x as x approaches zero?

I said zero, because sin(0) = 0. I graphed the function on my TI-84, and tested it with x values (10^-13) and -(10^-13) and it seemed to look pretty good, but according to my teacher it's 3/2. So if anyone could take a few minutes to look at it, I'd really appreciate it. Thanks.
I hope this helps:

So if you got something like this:

sin (x)
------- you can eliminate the "n" and get si(x) which is = 6.
n
the equation as it is written is undefined, because you cant divide by 0 (lol). So, you use L'Hopital's rule and get 3cos3x/2. Then, with direct substitution, cos(0)=1, and you get 3/2. Im pretty sure that's what your teacher was talking about.
Quote by Daneeka
I hope this helps:

So if you got something like this:

sin (x)
------- you can eliminate the "n" and get si(x) which is = 6.
n

The picture's funnier.
its 3/2

you factor out the sinx/x which leaves(sin x\x) (3/2). sin x/x =1 and 1 * (3/2) = 3/2
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theres some rule where lim x>0 (sin ax / bx) = a/b
You got to make the bottom = to theta in the top, so 3/2 *2x =3x
The lim x>0 for sinx/x=1 so you multiply that by the 3/2 and you just get 3/2. I might have explained this wrong, but I think it's the right idea.
(sin(3x))/2x

sin(3*0) = 0
2*0 = 0

now usually when you're dividing by zero its a bad thing
but 0/0 is an indeterminate form, which isn't as bad -> you can apply l'hopitals rule

lim (sin(3x))/2x = lim (sin(3x))'/(2x)' = lim (3*cos(3x)/2) = 3*cos(0)/2 = 3/2
It's 3/2. I use a simple formula for it. lim x->0 sin(x)/x is always 1. So, for sin(abc x) to be zero, we must have (abc x) in the denominater, where (abc x) is any value other than x. In this problem, move the two to before the lim which gives you 1/2 lim x->0 sin3x/x. Now, just multiply and divide by 3. Use the 3 in the denominator to match your denominator to the sine angle. Finally, you end up with 3/2 which is your answer.
Okay, thanks, the thing is, we haven't gone over L'Hospital's rule, even though I'm aware of what it is. We had our regular teacher quit last week, and the new one is awful, giving us a quiz on something we haven't learned.

It just really confused me, because when I graphed the function, it clearly approached zero from either side, and that's what a limit is, right?
Last edited by bananaboy at Oct 21, 2008,
u use a rule called hopital's rule.

u differentiate top and bottom

limit sin3x/2x = limit 3cos 3x/2

then u sub in x=0 and u get 3/2
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Quote by lanzaa
its 3/2

you factor out the sinx/x which leaves(sin x\x) (3/2). sin x/x =1 and 1 * (3/2) = 3/2

sin(a*b) =/= sin(a)*sin(b)
Since your lim x->0 makes it 0/0, Use hopital rule which is basically differentiating your numerator and denominator.

Makes it f(x)=sin3x/2x so differentiated function, f'(x)= 3 cos3x/2

Now make x->0

You'll get 3 cos 0 /2. Since cos 0 is 1, the answer is 3/2.
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