$\lim_{x \to x_0}{\frac{\arcsin x-\arcsin x_0}{x-x_0}}=?$

Question

I have the following problem:

$$ \lim_{x \to x_0}{\frac{\arcsin x-\arcsin x_0}{x-x_0}}=\text{?} $$

What I have:

Let $x=\sin t$. Then the problem becomes:

$$ \lim_{t \to t_0}{\frac{t-t_0}{\sin t-\sin t_0}} = \lim_{t \to t_0}{\frac{\frac{t-t_0}{2}}{\sin (\frac{t-t_0}{2}) \cos{\frac{t+t_0}{2}}}}=\lim_{t \to t_0}{\frac{1}{\cos(\frac{t+t_0}{2})}}=\frac{1}{\sqrt{1-\sin^2(t_0)}} $$

So I'm kind of stuck here. I'm not very good with trigonometric formulas. Assuming so far the solution is correct, how do I continue from here?

Thanks in advance!

Answer 1

$$\sin2C-\sin2D=2\sin(C-D)\cos(C+D)$$

Now $\lim_{t\to t_0}\cos\dfrac{t+t_0}2=\cos\dfrac{t_0+t_0}2$

See also: question about the limit $\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}$

Answer 2

It is the very definition of derivative of the given sine inverse function:

$$ \lim_{x \to x_0}{\frac{\arcsin x-\arcsin x_0}{x-x_0}} $$

$$ = \frac{d(arcsin x)}{dx}= \frac{1}{\sqrt {1-x^2}}$$

$$ =\frac{\pm 1}{\sqrt {1-x_0^2}},\,$$

EDIT1:

evaluated at $x=x_0.$ Double sign is applicable, sign is taken positive if in red portion, negative for blue portion of $f(x)$ and infinite slope at $x=\pm 1$ As inverse trig function is periodic and unbounded, only some portions are shown. Apologies, the bottom most arcsin function $(f< \pi/2)$ is incorrectly colored red, should be blue.