How to find $\lim_{x \to 0} ( \frac{1}{\sin(x)}- \frac{1}{\arcsin(x)})$

I want to do the problem without using L’Hopitals rule, I have
$$\frac{1}{\sin(x)}- \frac{1}{\arcsin(x)} = \frac{x}{\sin(x)}\frac{x}{\arcsin(x)}\frac{\sin(x)-\arcsin(x)}{x^2}$$
and I’m not quite sure about how to deal with the $\dfrac{\sin(x)-\arcsin(x)}{x^2}$, apparently its limit is $0$? In which case the whole limit would be $0$. But how would I show this without using l’Hopitals rule. Thanks for any help.

Solutions Collecting From Web of "How to find $\lim_{x \to 0} ( \frac{1}{\sin(x)}- \frac{1}{\arcsin(x)})$"

Using Taylor’s theorem:
$$\color{blue}{\sin x=x-x^3/6+x^5/120+O(x^7)}$$
$$\color{red}{\arcsin x=x+x^3/6+3x^5/40+O(x^7)}$$
So:
$$\begin{align}\left( \frac{1}{\color{blue}{\sin(x)}}- \frac{1}{\color{red}{\arcsin(x)}}\right)&=\frac{\color{red}{\arcsin x}-\color{blue}{\sin x} }{\color{blue}{\sin x}\arcsin x}\\&=\frac{\color{blue}{\color{red}{(x+x^3/6+O(x^5))}-(x-x^3/6+O(x^5))}}{\color{blue}{(x-x^3/6+O(x^5))}\color{red}{(x+x^3/6+O(x^5))}}\\&=\frac{+x^3/3+O(x^5)}{x^2+O(x^6)}\end{align}$$
So:
$$\lim \limits_{x \to 0}\left( \frac{1}{\sin(x)}- \frac{1}{\arcsin(x)}\right)=0$$

When $x\rightarrow0$

  • $\sin x\sim x-\frac{x^3}{6}$
  • $\arcsin x\sim x+\frac{x^3}{6}$
  • $\sin x-\arcsin x \sim (x-\frac{x^3}{6})-(x+\frac{x^3}{6})=-\frac{x^3}{3}$

$$\displaystyle \lim_{x\rightarrow 0}\frac{\sin x -\arcsin x}{\sin x \cdot \arcsin x }=\lim_{x\rightarrow 0}\frac{-\frac{x^3}{3}}{\sin x \cdot \arcsin x }=\lim_{x\rightarrow 0}\frac{-\frac{x^3}{3}}{ x ^2 }=\lim_{x\rightarrow 0}\frac{\frac{-x}{3}}{ 1 }=0\\
$$

Let $x \to 0^{+}$ and then we know that $\cos x < \dfrac{\sin x}{x} < 1$ (this is pretty standard from geometric arguments commonly used in proof of $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$ see this answer) and then we get $$\cos x – 1 < \frac{\sin x – x}{x} < 0$$ which further means that $$\frac{\cos x – 1}{x} < \frac{\sin x – x}{x^{2}} < 0\tag{1}$$ Now we can see that $$\frac{\cos x – 1}{x} = -\frac{2\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x} \to 0 $$ as $x \to 0^{+}$. Hence by using squeeze theorem on $(1)$ we get $$\lim_{x \to 0^{+}}\frac{\sin x – x}{x^{2}} = 0$$ and hence $$\begin{aligned}\lim_{x \to 0^{+}}\frac{\arcsin x – x}{x^{2}} &= \lim_{x \to 0^{+}}\frac{\arcsin x – x}{(\arcsin x)^{2}}\cdot\frac{(\arcsin x)^{2}}{x^{2}}\\
&= \lim_{t \to 0^{+}}\frac{t – \sin t}{t^{2}}\cdot\frac{t^{2}}{\sin^{2}t}\text{ (}\arcsin x = t)\\
&= 0\end{aligned}$$ It is now clear that $$\lim_{x \to 0^{+}}\frac{\sin x – \arcsin x}{x^{2}} = \lim_{x \to 0^{+}}\frac{\sin x – x}{x^{2}} – \frac{\arcsin x – x}{x^{2}} = 0$$ If $x \to 0^{-}$ then we can put $x = -y$ and $y \to 0^{+}$ and again the limit is $0$.

This is a nice example which shows that most usual limit problems don’t need advanced tools like L’Hospital’s Rule and Taylor series.