Question

Find the limit of sin(x)/x algebraically

Answer 1

um, the only ways I know are geometrical and using l'Hospital

Answer 2

I don't think a direct algebraic proof exists

Answer 3

the question is rather vague

Answer 4

other would be Taylor's expansion

Answer 5

oh yeah^

Answer 6

Ill look up l'Hospital then. I think i've heard it before. Thanks.

Answer 7

The Taylor series way is good too!

Answer 8

Its only been a few days in my Calc class, I dont think we've learned that rule yet.

Answer 9

when you learn the Taloyr expansion of sin and cos lim sinx/x and lim(1-cosx)/x are easy, but until then l'Hospiatal

Answer 10

taylor*

Answer 11

Got it, thanks!.

Answer 12

I have an idea, hold on a sec...

Answer 13

This is a long second. :)

Answer 14

latex takes along time to type

Answer 15

or the idea didn't pan out, eh?

Answer 16

It's not looking too promising, but since I don't have a pencil nearby I'm working it out in LaTeX... worst of both worlds ;) If you'd like to continue the discussion feel free...

Answer 17

Its fine. Ill ask some friends if I still dont get it. Thanks for all the help both of you.

Answer 18

I think we came to the conclusion that there are 3 ways:
Geometrical argument, which is a bit long-winded and difficult to do here
l'Hospital's rule (almost like cheating I feel sometimes)
and
Taylor expansion (the asker does not know Taylor series yet)
good luck!

Answer 19

Thanks, ill need it. :O

Answer 20

I would agree with that assessment, though I will keep toying around with finding a way to express it algebraically. Good luck.

Answer 21

\[ {\sin x \over x} = {{e^{i2x} - 1} \over (2 i x)e^{ix}}\]
using the property of e, you would get same result. though I'm not quite sure.

Answer 22

oh, how about trying the squeeze theorem?\[x<\tan x\implies x\cos x<\sin x\]so\[x\cos x<\sin x<x\implies\cos x<\frac{\sin x}x<1\]\[\lim_{x\to0}\cos x=1\]\[\therefore\lim_{x\to0}\frac{\sin x}x=1\]