Answer
$1$
Work Step by Step
Multiply the limit with $\displaystyle \frac{x^{2}}{x^{2}}$ and rewrite:
$L=\displaystyle \lim_{x\rightarrow 0^{+}}\frac{\sin^{-1}x^{2}}{x^{2}}\cdot\lim_{x\rightarrow 0^{+}}\frac{x^{2}}{(\sin^{-1}x)^{2}}=L_{1}\cdot L_{2}$
$L_{1}=\displaystyle \lim_{x\rightarrow 0^{+}}\frac{\sin^{-1}x^{2}}{x^{2}}\qquad $... substitute $\left[\begin{array}{l}
t=x^{2}\\
t\rightarrow 0^{+}
\end{array}\right]$
$L_{1}=\displaystyle \lim_{t\rightarrow 0^{+}}\frac{\sin^{-1}t}{t}\qquad $... $\displaystyle \frac{0}{0}$, apply L'Hopital's rule
See table 7-4,$ \displaystyle \quad \frac{d(\sin^{-1}u)}{dx}=\frac{1}{\sqrt{1-u^{2}}}\frac{du}{dx}, |u| \lt 1$
$L_{1}=\displaystyle \lim_{t\rightarrow 0^{+}}\frac{\frac{1}{\sqrt{1-t^{2}}}}{1}=\lim_{t\rightarrow 0^{+}}\frac{1}{\sqrt{1-t^{2}}}=1$
$L_{2}=\displaystyle \lim_{x\rightarrow 0^{+}}\frac{x^{2}}{(\sin^{-1}x)^{2}}=[\lim_{x\rightarrow 0^{+}}\frac{x}{\sin^{-1}x}]^{2}$
The limit has the form $\displaystyle \frac{0}{0}$; apply L'Hopital's rule
$L_{2}= [\displaystyle \lim_{x\rightarrow 0^{+}} \displaystyle \frac{1}{\frac{1}{\sqrt{1-x^{2}}}}]^{2}=[\lim_{x\rightarrow 0^{+}}\sqrt{1-x^{2}}=1]^{2}$
$L_{1}\cdot L_{2}=1\cdot 1=1$
