Question

show that if x is in degrees, then lim [(sin x)/x] = (pi / 180) as x -> 0

Answer 1

There's no easy way to show the (sinx/x) part really As x goes to 0, (sinx/x) goes to 1, but other than remembering that, I don't believe there's an easy way to show that. The proof involves sections and triangles cut out of a circle, so I think it's kind of a bit out there to have students know it.
If absolute proof is needed, Ican try and show ya, but I'm not familiar with it either. But apart from saying that (sinx/x) is 1 as limit x goes to 0 is that pi is the same as 180 degrees, so it's essentially division by itself which gives 1.
Maybe @pgpilot326 knows a good way to show that (sinx/x) as x ->0 = 1

Answer 2

I mean, how are you supposed to show it? We can see that if we plug it in we get 0/0 which is an indeterminate form so then we can use L'Hopital's rule to take the limit of the ratio of the derivative of the top over the derivative of the bottom:

\[\lim_{x \rightarrow 0}\frac{ sinx }{ x }=\lim_{x \rightarrow 0}\frac{ \frac{ d }{ dx} (sinx) }{\frac{ d }{ dx} ( x) }=\lim_{x \rightarrow 0}\frac{ cosx }{ 1 }=1\]

Answer 3

Guess it depends on whether or not they've made it to l'hopitals rule or not. I learned the sinx/x thing way way before l'hopitals rule. It was just something that was supposed to be memorized with an obscure proof, lol.

Answer 4

the proof isn't so obscure. in fact it's crucial in establishing the derivatives of the trig functions. but that's not really the issue here. it's dealing with the degree issue.

Answer 5

Well, it doesn't seem like a proof that would be expected for students to know. A lot of the things like that just seem to be things that need to be assumed or simply known. i suppose that's what I mean by obscure.

Answer 6

Sure we can do the squeeze theorem aswell. Ok so looking at sinx we can see that:

\[\frac{ 1 }{ 20 }x^2\ge sinx \ge20x^2\] on the interval of importance.

Now divide everything by x and take the limits. Tah dah the limit is 0.

Answer 7

lol