Question

Can someone explain me why the following limit is equal to 1 w/o using L'Hospital 's rule?

Answer 1

\[lin_{h \rightarrow 0} {\sin(h) \over h} = 1\]

Answer 2

it's an identity.. and the only way to prove it is l'hopital.. i may be mistaken but thats the only way that i know how to prove that..

Answer 3

as sin(h) gets really small; its about equal to "h"; and the smaller it gets the closer to 1 it gets

Answer 4

http://www.khanacademy.org/video/proof--lim--sin-x--x?playlist=Calculus

Answer 5

type in sin(.00000001)/.00000001 and see what the calc gives yah

Answer 6

I get the intuitive idea, but is there any algebraic way?

Answer 7

that is the algeraic way i think; it just is what it is :)

Answer 8

you have an maybe if we express h in terms of radians? or is it already in terms of radians?

Answer 9

you have to use something called the "Squeeze theorem"

Answer 10

http://www.youtube.com/watch?v=Ve99biD1KtA

Answer 11

@amistre the only way this is true is if h is in "radians". as a function, sine is a function of numbers, not of "degrees" or "radians". just like
\[f(x)=x^2+1\] is a function of numbers. as such, it corresponds to the usual trig ratios of sides of triangles only if the angles are measured in radians

Answer 12

and in fact, if you are using "degrees" as your input, the derivative of sine is NOT cosine. really.

Answer 13

the basic geometric argument boils down to saying that
\[\cos(x)<\frac{sin(x)}{x}<1\] near 0, and then as
\[x->0\] the left hand side goes to 1 as well and therefore the limit is 1. but if you want a simple snap proof WITHOUT using derivatives there is not one.

Answer 14

if you've had anlayss consider a sequence \[(x_n) \rightarrow 0 \] and use the continuity proof for \[(f(x_n)) \rightarrow f(x)\] and show that converges to 1 using a conventional \[\epsilon - \delta\] proof . If your more versed with topology you could define the filter converging to 0 and apply the function to it in a similar manner.

Answer 15

with the filter showing it converges to1

Answer 16

Thanks a lot. That was one of the most clever proofs that I have ever seen. It was beautiful, too.