How can I compute this limit without L'Hôpital's rule: $x$ approaches zero for $\arcsin(x^2)/x^2$?

Question

I think I can evaluate it with a substitution but not sure about the passages.

Answer 1

In THIS ANSWER, I showed using standard, non-calculus-based tools that the arcsine function satisfies the inequalities

$$x\le \arcsin x\le \frac{x}{\sqrt{1-x^2}}$$

Therefore, we have

$$x^2 \le \arcsin(x^2)\le \frac{x^2}{\sqrt{1-x^4}}$$

whereupon dividing both sides by $x^2$ yields

$$1 \le \frac{\arcsin(x^2)}{x^2}\le \frac{1}{\sqrt{1-x^4}}$$

Now applying the squeeze theorem, we obtain

$$\lim_{x\to 0}\frac{\arcsin(x^2)}{x^2}=1$$

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Alternatively, we could have enforced the substitution $x^2=\sin y$. Therefore

$$\frac{\arcsin(x^2)}{x^2}=\frac{y}{\sin y}$$

and since $\lim_{x\to 0}y=0$, we have

$$\lim_{x\to 0}\frac{\arcsin(x^2)}{x^2}=\lim_{y\to 0}\frac{y}{\sin y}=1$$