Question

I'm having trouble with this calc problem.
The lim as x approaches 0=sin^2(4x)/(6x)
I'm not seeing a way to get the bottom to not=0.

Answer 1

\[\huge \lim_{x \rightarrow 0}\frac{\sin^2(4x)}{6x}\]
This limit?

Answer 2

Yes that is the limit.

Answer 3

and for this, you need to know what lim x->0 sin x/x equals....

Answer 4

Too bad you probably haven't heard of, or aren't allowed to use L'hopital's rule, or you'll find this to be ridiculously easy :D

@hartnn has got the main point of this, you know what that limit is?
\[\huge \lim_{x \rightarrow 0}\frac{\sin \ x}{x}\]

Answer 5

That equals 1 correct? Because don't you just take derivative of numerator and denominator and divide them to get cos(x)/1 and plugging in 0 we get 1/1?

Answer 6

You took the derivative of the numerator and the denominator? But that's L'Hopital's rule!
So you're allowed to use that rule after all? :D

Answer 7

\[\frac{sin^24x}{6x}=\sin 4x\cdot\frac{\sin 4x}{6x}\]

Answer 8

even if its allowed, why to use it, when we can solve without it..

Answer 9

I've never heard of the rule. I was just told the answer to sin(x)/x in high school and still remember it.

Answer 10

Oh... okay, well, follow @sirm3d 's lead, then...
\[\huge \frac{\sin^24x}{6x}=\sin 4x\cdot\frac{\sin 4x}{6x}=\frac{2\sin x}{3}\cdot\frac{\sin \ 4x}{4x}\]

Answer 11

The limit of the product is the product of their limits... IF both limits exist :)

Answer 12

I'm confused about where I'm supposed to be going with what information you just gave me.

Answer 13

\(\huge \frac{sin^24x}{6x}=\sin 4x\cdot\frac{\sin 4x}{6x}=\dfrac{4\sin 4x}{6}\cdot\frac{\sin 4x}{4x}\)
you know what will be the limit of 2nd term ?

Answer 14

the limit of that is 1 right?

Answer 15

yes :)
and 1st term ?

Answer 16

just plug in x=0

Answer 17

^Because sin is continuous :)

Answer 18

If you plug in 0 for the first one you'll get 0

Answer 19

That's right :)
The limit of the product is the product of the limits IF both limits exist....
both limits exist... therefore...? ;)

Answer 20

So the answer for the limit is 0?

Answer 21

Nailed it... XD

Answer 22

good :)

Answer 23

the point here is to uunderstand the adjustments so that we could use standard formula of sin x/x

Answer 24

I see. So not using L'Hopital's how do we know that sin(x)/x=1

Answer 25

oh, you need the proof of that standard formula ?

Answer 26

I've never been taught why that is. I was told that 2 years ago in my junior year of high school. In my pre-calc class. I was never given the reason why that is.

Answer 27

It's quite long, though... and involves drawing the triangle and the unit circle, etc...
It might even be shorter to prove L'Hopital first...

Answer 28

yeah, if you really interested ,do some reading...
http://www.proofwiki.org/wiki/Limit_of_Sine_of_X_over_X

Answer 29

No, wait... anything but that... been there, so many complications...
I COULD try to prove it here, but that takes some patience from you, @kiszer4148 ...
I proceed only at your signal ;)

Answer 30

Haha I'm very interested in knowing. But only if you have the time. I can tell you I don't have class for another 4 hours.

Answer 31

I... am home, from school :)
Let's draw the unit circle....

Answer 32

Now, for the purposes of this proof, let's assume for the moment that x takes a value from
0 to (1/2)π

Answer 33

Now let's draw these lines...
So, we let x be this arclength, and since this is the unit circle, this x is indeed between 0 and a half π

Answer 34

This arclength, sorry...

Answer 35

Are you getting it so far?

Answer 36

First, notice that this point here...
And this point here