How to evaluate the limit $\lim_{x\to0} \frac{\sin^{-1}{x}}{x}$ without using L Hospital I need to evaluate this limit without using L Hospital. This is what I have done so far:
Let's begin with the assumption $x = \sin{t}$
Then as $x\to0$, $t\to0$
Now substituting $\sin{t}$ in place of $x$ in the limit
$$
\begin{align*}
    \lim_{x\to0} \frac{\sin^{-1}{x}}{x} &= \lim_{t\to0} \frac{\sin^{-1}{\sin{t}}}{\sin{t}} \\
    &= \lim_{t\to0} \frac{t}{\sin{t}} \\
    &= \frac{1}{\lim_{t\to0} \frac{\sin{t}}{t}} \\
    &= \frac{1}{1} = 1
\end{align*}
$$
What I am concerned about is that by the assumption $x=\sin{t}$, $x$ is being limited to the interval $[-1,1]$. But as we are concerned about the limit when $x\to0$, I should not worry about the limitation? Besides the domain of $\sin^{-1}{x}$ is $[-1,1]$.
So, are there any errors in the above approach?
Thanks
 A: We need to use here the Taylor series:
$$
\bbox[lightgreen]
{
\arcsin(x)=\left(x+\frac{x^{3}}{6}+\frac{3x^{5}}{40}+....\right)
},\tag{1}
$$
Therefore,
$$
\lim_{x\to{0}}\frac{\arcsin{x}}{x}=\lim_{x\to{0}}\frac{\left(x+\frac{x^{3}}{6}+\frac{3x^{5}}{40}+....\right)}{x}=
\\
=\lim_{x\to{0}}\frac{x\left(1+\frac{x^{2}}{6}+\frac{3x^{4}}{40}+....\right)}{x}=\lim_{x\to{0}}\left(1+\frac{x^{2}}{6}+\frac{3x^{4}}{40}+....\right)=1.
$$
Here is the result:
$$
\bbox[lightblue]
{
\lim_{x\to{0}}\frac{\arcsin{x}}{x}=1
}.
$$
A: This is correct. The only thing I would do differently is start with what's known to be true, i.e. the limit
$$\lim_{t\to 0}\frac{t}{\sin t}=1 $$
then substitute $t=\sin^{-1}x\to 0$ as $x\to 0$ to arrive at the limit that needs to be shown. As for your question about limiting the values of $x$, since we're talking about a limit as $x\to 0$, you may assume that $x\in(-\varepsilon,0)\cup(0,\varepsilon)$ for an arbitrary $\varepsilon>0$. This simply follows from the definition of a limit.
