Answer
$1$
Work Step by Step
Rewrite the limit as the product of two limits
$L= \displaystyle \lim_{x\rightarrow 0}\frac{\tan^{-1}x^{2}}{x^{2}}\cdot\lim_{x\rightarrow 0}\frac{x}{\sin^{-1}x}=L_{1}\cdot L_{2}$
$L_{1}= \displaystyle \lim_{x\rightarrow 0}\frac{\tan^{-1}x^{2}}{x^{2}} \qquad $is of the form $\displaystyle \frac{0}{0}$,
but it can be further simplified before differentiating (applying L'Hospital's rule).
Introduce a new variable $t=x^{2}$. When $x\rightarrow 0$, we also have $t\rightarrow 0$
$L_{1}= \displaystyle \lim_{t\rightarrow 0}\frac{\tan^{-1}t}{t}\qquad $
Now, apply the rule:
From table 7-4:$\quad \displaystyle \frac{d(\tan^{-1}u)}{dx}=\frac{1}{1+u^{2}}\frac{du}{dx}$
$L_{1}= \displaystyle \lim_{t\rightarrow 0}\frac{\frac{1}{1+t}}{1}= \displaystyle \lim_{t\rightarrow 0}\frac{1}{1+t}=\frac{1}{1+0}=1$
$L_{2}=\displaystyle \lim_{x\rightarrow 0}\frac{x}{\sin^{-1}x}\qquad $
Apply L'Hospital's rule, with $\displaystyle \frac{d(\sin^{-1}u)}{dx}=\frac{1}{\sqrt{1-u^{2}}}\frac{du}{dx}$
$L_{2}=\displaystyle \lim_{x\rightarrow 0}\frac{1}{\frac{1}{\sqrt{1-x^{2}}}}=\lim_{x\rightarrow 0}\sqrt{1-x^{2}}=\sqrt{1-0}=1$
$L=L_{1}\cdot L_{2}=1$
