I fail to see how you can "see" finer than sin(h) -> 0 & cos(h) -> 1
From the limit definitions you actually need :
* (1-cos(h)) / h -> 0
* sin(h)/h -> 1
(which correspond to the derivatives at 0).
If you zoom in sufficiently at x = 0, f(x) = sin(x) looks indistinguishable from f(x) = x, whereas g(x) = cos(x) looks indistinguishable from g(x) = 1.
(also, sin(x) is negative approaching 0 from the left and positive approaching 0 from the right)
You can't use the plot as you only know the triangle definition yet. (And "looks indistinguidable" is rather handwavy).
> (also, sin(x) is negative approaching 0 from the left and positive approaching 0 from the right)
That only tells you that sin(0)=0
Curvature is inverse of radius.
Decreasing angle is equivalent to increasing radius, and this decreasing curvature. This, as angle decreases, the curve becomes close to a straight line, and that straight line approaches a vertical line.
As usual, 3B1B created a quintessential visualization and explanation.https://m.youtube.com/watch?v=S0_qX4VJhMQ
I perfectly understand that around 0, sin(x) ~ x and cos(x) = 1 + o(x) but it isn't obvious geometrically, unlike what the article implies.
From my point of view, increasing radius / decreasing curvature only gets you sin(x) -> 0 ; cos(x) -> 1, but that isn't enough to obtain the derivatives.
I found a geometric proof in [1] but that part is the longest and hardest of the page. I was wondering whether the author found a clearer way to express is.
[1] https://www.mathsisfun.com/calculus/derivatives-trig-proof.h...
EDIT: after looking at 3B1B's video, the "small" triangle d(sinΘ) by dΘ figure would be a better way to explain the derivative, rather than an "not hard to see geometrically" approximation that isn't enough to conclude.