Find the limit of $\lim_{ x\to 0 } \sin^{-1}(x) /\sin(3x)$ without using L'Hopital's rule

Question

I'm new to calculus and I'm not sure how to deal with the inverse $\sin$ here. $$ \lim_{x \to 0} \frac{ \sin^{-1}(x) }{\sin(3x)} $$

Answer 1

Hint

$$\lim_{x\to 0} \dfrac{\arcsin x}{x}=\lim_{x\to 0} \dfrac{x}{\sin x}=1.$$

Answer 2

With substitution $x\to\sin t$ you have $$\lim_{x\to 0}\frac{\sin^{-1}x}{\sin3x}=\lim_{t\to 0}\frac{t}{\sin(3\sin t)}=\lim_{t\to 0}\frac{3\sin t}{\sin(3\sin t)}\cdot\frac{t}{3\sin t}=\color{blue}{\dfrac13}$$

Answer 3

$\sin ^{-1}x= t$

$\lim\limits_{x→0}\sin 3t→ 3t$

Therefore:

$\lim\limits_{x→0} \dfrac{\sin^{-1} x}{\sin 3x}=\dfrac{t}{3t}=\dfrac{1}{3}$

Answer 4

$$\lim \limits_{ x\to 0 } \frac { { \sin }^{ -1 }(x) }{\sin{(3x)} }=\\$$ We substitute $x$ by $\sin (t)$. This is possible because $\sin$ is a strict increasing continous function near $0$ and $sin(0)=0$ and get $$=\lim \limits_{ t\to 0 } \frac { { \sin }^{ -1 }(\sin(t) )}{\sin{(3\sin(t))} }=\\$$ $$=\lim \limits_{ t\to 0 } \frac {t}{\sin{(3\sin(t))} }= $$ $$=\lim \limits_{ t\to 0 } \frac {t}{\sin(t)}\frac {\sin(t)}{\sin{(3\sin(t))} }= $$ $$=\lim \limits_{ t\to 0 } \frac {t}{\sin(t)}\lim \limits_{ t\to 0 }\frac {\sin(t)}{\sin{(3\sin(t))} }= $$ $$=\lim \limits_{ t\to 0 } \frac {t}{\sin(t)}\frac{1}{3}\lim \limits_{ t\to 0 }\frac {3\sin(t)}{\sin{(3\sin(t))} }= $$ and if we substitute $3 \sin t$ by $u$ then we get $$=\lim \limits_{ t\to 0 } \frac {t}{\sin(t)}\frac{1}{3}\lim \limits_{ u\to 0 }\frac {u}{\sin{(u)} }= $$ $$=\frac{1}{3}$$

because $$\lim \limits_{ y\to 0 } \frac {y}{\sin(y)}=1$$