Spivak: $f(x)=\sqrt{1-x^2}$ on $[-1,1]$. $\mathscr{L}(x)$ defined as length of $f$ on $[x,1]$. Show $\mathscr{L}(x)=\int_x^1 \frac{1}{\sqrt{1-t^2}}dt$ The following problem is from Ch. 15 of Spivak's Calculus. I show my attempted solution with a question, and then try to interpret the solution from the solution manual, with another question about that.

*28. This problem gives a treatment of the trigonometric functions in terms of length, and uses Problem 13-25. Let $f(x)=\sqrt{1-x^2}$ for
$-1\leq x\leq 1$.
Define $\mathscr{L}(x)$ to be the length of $f$ on $[x,1]$.
(a) Show that
$$\mathscr{L}(x)=\int_x^1 \frac{1}{\sqrt{1-t^2}}dt\tag{1}$$

Here is my attempt at a solution
In Problem 13-25 we are asked to show that the length of a function $f$ on $[a,b]$ is $\int_a^b \sqrt{1+(f'(t))^2}dt$ if the integrand is integrable on $[a,b]$.
For our $f$ we have
$$f'(x)=\frac{-x}{\sqrt{1-x^2}}$$
$$\sqrt{1+(f'(x))^2}=\frac{1}{\sqrt{1-x^2}}$$
Therefore, if $\frac{1}{\sqrt{1-x^2}}$ is integrable on $[x,1]$ then $(1)$ is the length of $f$ on $[x,1]$.
The integral in $(1)$ is improper since $\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$ is unbounded near $1$.
I believe this improper integral can be defined as
$$\lim\limits_{\epsilon\to 1^-} \int_x^{\epsilon} \arcsin'(t)dt$$
If we compute this limit, is that sufficient to prove that $(1)$ is true?
Spivak's solution manual does something different

length of $f$ on $[x,1-\epsilon]$
$$=\int_x^{1-\epsilon} \sqrt{1+[f'(t)]^2}dt$$
$$= \int_x^{1-\epsilon} \frac{1}{\sqrt{1-t^2}}dt\tag{2}$$
To obtain the desired expression for $\mathscr{L}(x)$ we must then use
the fact that
$$\lim\limits_{\epsilon\to 0}\ (\text{length of }f\text{ on }
 [x,1-\epsilon])=\text{ length of }f\text{ on }[x,1]\tag{3}$$
This is proved as follows. First of all, the following figure shows
that the "length of $f$ on $[x,1]$" does make sense; in fact, the
length of $f$ on $[0,1]$ is $\leq 2$.

The same sort of figure also shows that the length of $f$ on
$[1-\epsilon,1]$ is $\leq 2\epsilon$. The desired limit then follows
from this inequality and the fact that
$$\text{length of }f\text{ on } [x,1] = \text{ length of }f \text{ on
 } [x,1-\epsilon]+\text{ length of }f\text{ on } [1-\epsilon,1]$$
The proof of this latter fact is very similar to the corresponding
assertion for integrals.

Let me try to interpret the assertions perhaps slightly more rigorously I hope.
The length of $f$ on $[x,1-\epsilon]$ is just a regular (ie, not improper) integral $\int_x^{1-\epsilon} \frac{1}{\sqrt{1-t^2}}dt$.
The length of $f$ on $[x,1]$ is the improper integral $\int_x^1 \frac{1}{\sqrt{1-t^2}}dt$.
We can write this integral as $$\int_x^1 \frac{1}{\sqrt{1-t^2}}dt=\int_x^{1-\epsilon} \frac{1}{\sqrt{1-t^2}}dt+\int_{1-\epsilon}^1 \frac{1}{\sqrt{1-t^2}}dt\tag{4}$$
where $\int_{1-\epsilon}^1 \frac{1}{\sqrt{1-t^2}}dt$ is the improper integral. But with a geometric argument we can see that this integral is both larger than $0$ and $<2\epsilon$
Now we take the limit of $(4)$ at $\epsilon \to 0$ we have
$$\int_x^1 \frac{1}{\sqrt{1-t^2}}dt=\lim\limits_{\epsilon\to 0} \left [ \int_x^{1-\epsilon} \frac{1}{\sqrt{1-t^2}}dt+\int_{1-\epsilon}^1 \frac{1}{\sqrt{1-t^2}}dt\right ]\tag{5}$$
$$=\lim\limits_{\epsilon\to 0} \int_x^{1-\epsilon} \frac{1}{\sqrt{1-t^2}}dt\tag{6}$$
At this point, what have we shown exactly? Don't we still have to show that this limit exists (by computing it in a simple way using $\arcsin$)?
 A: You are right: We must check that the limit $(6)$ exists. However, in your presentation you introduce the improper integral $\int_{1-\epsilon}^1 \frac{1}{\sqrt{1-t^2}}dt$. This does not make sense unless you know that it exists. And even if you know that, it requires an argument that this integral is the length of $f$ on $[1-\epsilon,1]$. So let us go back.
For a continuous $f :[a,b] \to \mathbb R$ an a partition $P = (t_0,\ldots,t_n)$ of $[a,b]$ Spivak defines
$$\ell(f,P) =\sum_{i=1}^n \sqrt{(t_i-t_{i-1})^2 + (f(t_i) - f(t_{i-1}))^2} $$
If there exists an upper bound of all $\ell(f,P)$ for all partitions $P$, then the least upper bound is called the length of $f$ which we shall denote by $\ell(f)$. In an exercise Spivak asks to prove that if $f$ is differentiable and $f'$ is integrable, then $\ell(f)$ exists and is given by
$$\ell(f) = \int_a^b \sqrt{1 + f'(t)^2} dt$$
The problem here is that for $f(x) = \sqrt{1-x^2}$ no derivation exists at $x = \pm1$ and moreover $f'(x)$ is not bounded on $(-1,1)$. Only on closed subintervals $[a,b] \subset (-1,1)$ we can say that $\ell(f \mid_{[a,b]})  = \int_a^b \frac{1}{\sqrt{1-t^2}}dt$.
Thus we have to check first that the $\ell(f\mid_{[x,1]},P)$ are bounded above. Spivak gives a "graphical proof" (which can be easily formalized) that for $x \ge 0$ we always have
$$\ell(f \mid_{[x,1]},P) \le 2(1-x) . \tag{1}$$
One can easily check that $(1)$ also holds for $x \le 0$. Thus $\mathscr L(x)$ is well-defined for all $x \in [-1,1]$ with $\mathscr L(x) \le 2(1-x)$. It remains to compute $\mathscr L(x)$. To do so, we use the fact that for $x > -1$ we have
$$\mathscr L(x) = \ell(f \mid_{[x,1-\epsilon]}) + \mathscr L(1-\epsilon) .$$
Therefore
$$\lvert \mathscr L(x) - \ell(f \mid_{[x,1-\epsilon]}) \rvert = \mathscr L(1-\epsilon) \le 2\epsilon . \tag{2}$$
But
$$ \ell(f \mid_{[x,1-\epsilon]}) = \int_x^{1-\epsilon}  \frac{1}{\sqrt{1-t^2}}dt = \arcsin(1-\epsilon) - \arcsin(x)\\ \to \arcsin(1) -  \arcsin(x) = \pi/2 - \arcsin(x) \text{ as } \epsilon \to 0.$$
Thus $(2)$ shows that
$$\mathscr L(x) = \lim_{\epsilon \to 0} \ell(f \mid_{[x,1-\epsilon]}) = \pi/2 - \arcsin(x).$$
The case $x=-1$ can be treated by similar arguments using $\mathscr L(-1) = \ell(f \mid_{[0,\epsilon]}) + \mathscr L(\epsilon)$.
