Why is $\lim_{x \to 0}\frac{\sin^{-1}x}{x} = \lim_{x \to 0}\frac{1/\sqrt{1-x^2}}{1}$? I'm a beginner in Calculus and I'm studying L'Hospital's Rule. 
I have to calculate the following limit: $$\lim_{x \to 0}\frac{\sin^{-1}x}{x}$$.
My solutions manual presents the following equation:
$$\lim_{x \to 0}\frac{\sin^{-1}x}{x} = \lim_{x \to 0}\frac{1/\sqrt{1-x^2}}{1}$$
While applying L'Hospital's Rule, I understand that the derivative of $x$ is $1$, but I can't figure out why the derivative of $1/\sin x$ would be $1/\sqrt{1-x^2}$. I would rather think this should be the result. 
Any explanations for this?
 A: Firstly, note that $\sin^{-1}(x)$ generally refers to what is often known as $\arcsin(x)$. From there, I think the only thing you are missing is why 
$$ \frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}.$$
I'll assume you are familiar with implicit differentiation. Set $\arcsin(x)=y$. Then $\sin(y)=x$. We have 
$$ \frac{d}{dx}\sin(y)=\frac{d}{dx}(x)$$
$$ \frac{dy}{dx}\cos(y)=1$$
$$ \frac{dy}{dx}=\frac{1}{\cos (y)}.$$
Now, $\cos^2y+\sin^2y=1$. Substituting, we have 
$$ \frac{dy}{dx}=\frac{1}{\sqrt{1-\sin^2 (y)}}.$$
Using the identities from the beginning, we have 
$$ \frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$$
as desired.
A: This is inverse notation:
$$f(f^{-1}(x))=f^{-1}(f(x))=x$$
If $f(x)$ is bijective/one-to-one.
By chain rule, when $x\in[-\pi,\pi]$:
$$x=\sin(\sin^{-1}(x))$$
$$1=\cos(\sin^{-1}(x))(\sin^{-1}(x))'$$
$$(\sin^{-1}(x))'=\frac1{\cos(\sin^{-1}(x))}\\=\frac1{\sqrt{1-(\sin(\sin^{-1}(x)))^2}}\\=\frac1{\sqrt{1-x^2}}$$
Where I use the Pythagorean theorem.

However, more easily, note that by letting $x=\sin u$, we get:
$$\lim_{x\to0}\frac{\sin^{-1}x}x=\lim_{u\to0}\frac{\sin^{-1}(\sin(u))}{\sin u}=\lim_{u\to0}\frac u{\sin u}=1$$
And by direct substitution, the second limit follows.
A: So $\sin ^{-1} x \ne (\sin x)^{-1}$, but it means inverse mapping, i.e. if $y= \sin(x)$, then we say $x=\sin^{-1} y$ 
Here, we need to be careful about the domain and range when defining those mappings and deriving the derivative, because $\sin x$ is not a bijection - as a matter of fact, we only define $\sin^{-1} x$ over $[-\frac{\pi}{2},\frac{\pi}{2}]$
Let $y=\sin^{-1}x: [-1,1] \mapsto [-\frac{\pi}{2},\frac{\pi}{2}]$, so $x=\sin y$. Thus
$$1=\cos y \cdot y'$$
$$y'=\frac{1}{\cos y}$$
But if $x=\sin y$, we have $\sqrt{1-x^2}=\cos y$, because $\cos y \ge0$ for $y \in [-\frac{\pi}{2},\frac{\pi}{2}]$
Thus
$$y'=\frac{1}{\sqrt{1-x^2}}$$
A: The notation $\sin^{-1}x$ is used to mean the inverse sine function, whereas $(\sin(x))^{-1}$ is used to mean $1/\sin x$.  To avoid ambiguity, $\arcsin x$ is often used for the inverse sine function.
A: $$\sin^{-1}(x) \neq \frac{1}{sin(x)}$$
but it is notation for the inverse function of the sine function. Another frequently used notation is $\arcsin(x)$
Because $$(\arcsin(x))'  = \frac{1}{\sqrt{1-x^2}}$$
the result follows. 
A: $ \sin^{-1} x $ is just a notational difference from $\arcsin x$, the inverse sine function, which represents $ \sin y = x \quad (x \in (-\pi/2, \pi/2))$. The derivative is given by,
$$ \frac{d}{dx} [\sin^{-1} x] = \frac{1}{\sqrt{1-x^2}} $$
So the limit simplifies to,
$$ \lim_{x \to 0} \frac{1}{\sqrt{1-x^2}} = 1 $$
