I have been asked to find the torsion of a helix (an exercise of a book) using the following definition. I've tried solving it no avail (attempting to find the angle directly). I would appreciate if someone could explain to me how the author solves it. In particular the third step (you'll see what I'm talking about). For those who want to look at the book here it is.
Definition. (Torsion) The torsion of a curve $x=x(t)$ at the point $s_0$ is
$$\frac{1}{T} = \pm \lim_{\Delta s_0 \rightarrow 0} \frac{\Delta \phi}{\Delta s}$$
where $\Delta \phi$ is the angle between the osculating planes at the points $s_0$ and $s_0 + \Delta s$.
Excercise
Find the torsion of the helix at $s = 0$,
$x1 = cos \frac{s}{\sqrt{2}}$
$x2 = sin \frac{s}{\sqrt{2}}$
$x3 = \frac{s}{\sqrt{2}}$
What I have so far
So taking the dot product of the normals of the osculating planes and dividing by the products of norms should give me:
$$cos \Delta \phi = \frac{1}{2} cos \frac{\Delta s}{\sqrt{2}} + \frac{1}{2}$$
That part is clear and that same step is in the book. However then, the author takes away 1 from each side, factors and divides by $\Delta s^2$, giving:
$$\frac{2(1 - cos \Delta \phi)}{\Delta s^2} = \frac{1-cos \frac{\Delta s}{\sqrt{2}}}{\Delta s^2}$$
Okay so far so good. But then the author claims that (which I don't understand how):
$$ \frac{1}{T^2} = \lim_{\Delta s \rightarrow 0} \frac{1 - cos \frac{\Delta s}{\sqrt{2}}}{\Delta s^2} = \frac{1}{4}$$
And so,
$$\frac{1}{T} = \pm \frac{1}{2}$$
I know how to solve the limit but I don't understand why we equalled $\frac{1}{T^2}$ like that.