Can anyone remember...

[Liokae]

... how to find the limit as x approaches 0 of sin(x)/x *without* using L'Hopital's rule?

[Diamond187]

I don't think there is a way, my first calc teacher just gave it to me as an identity.

[Afterburner]

Look at this:

sin x < x < tan x

(this is easy to prove, look elsewhere)

(< means less than or equal to, i can't find that symbol)

divide by sin x:

1< x/sin x < 1/cos x

reverse it:

1>sin x/x > cos x

Take the limit as x goes to zero. Cos x turns into 1, and thus sin x/x must also go to one because:

1> sin x/x > 1

QED.

[Diamond187]

That works, and here I was looking for some bizarre trig identity to make it happen.

[Liokae]

Hmm. *Does* that work, though? Because while sin(x) <= x is true as far as I can see, x <= tan(x) isn't.

[Diamond187]

no, it is true for positive x, you could look at the graphs, plot some points or just look at the derivatives, they should all hold.

[Afterburner]

The proof is a geometric one...use areas and stuff...and yes it's only for x>0. For x<0 you gotta use L'Hopital....

[Liokae]

It doesn't work for either negative *or* positive x.

If x = pi, tan(x) = 0. Pi > 0.

If x = -1.08, tan(x) = -1.87. -1.08 > -1.87.

[Diamond187]

Ok, so it works for 0<=x<pi/2, and that's all you need for the positive side ( I think).

[Liokae]

For a general proof, it has to apply everywhere; it doesn't. So we're back to L'Hoptals rule, or just calling it a limit identity.

[Afterburner]

No...of course we are limiting x to between 0 and pi/2, otherwise all sorts of weird things pop up. That's standard.

[Diamond187]

well, typically you limit it from 0 to 2pi because otherwise you end up repeating yourself. In other cases, tho, other values of theta are also ignore. In tanx's case, usually you go from -pi/2 to pi/2 since that's a full cycle, but it only works for 0 -> pi/2. My suggestion for a general proof, would be to look for ways to multiply sinx/x by "1" to force a trig identity which could help.