How to evaluate the limit $\lim_{x\to0} \frac{\sin^{-1}{x}}{x}$ without using L Hospital

Question

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

Answer 1

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.

Answer 2

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: $$ \bbox[lightblue] { \lim_{x\to{0}}\frac{\arcsin{x}}{x}=1 }. $$