Answer
$1$
Work Step by Step
$x$ is approaching 0 from the right, so it is positive $(x=\sqrt{x}\cdot\sqrt{x}$)
Rewrite the limit
$L=\displaystyle \lim_{x\rightarrow 0^{+}}\frac{(\tan^{-1}\sqrt{x})^{2}}{(\sqrt{x})^{2}}\cdot\lim_{x\rightarrow 0^{+}}\frac{1}{\sqrt{x+1}}=L_{1}\cdot L_{2}$
$L_{2}=1$ by direct substitution.
$ L_{1}=[\displaystyle \lim_{x\rightarrow 0^{+}}\frac{\tan^{-1}\sqrt{x}}{\sqrt{x}}]^{2}\qquad$
The form of the limit is $\displaystyle \frac{0}{0}$. Apply L'Hospital's rule.
From table 7-4, $\displaystyle \quad \frac{d(\tan^{-1}u)}{dx}=\frac{1}{1+u^{2}}\frac{du}{dx}$
$L_{1}=[\displaystyle \lim_{x\rightarrow 0^{+}}\frac{\frac{1}{1+(\sqrt{x})^{2}}\cdot\frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}]^{2}=[\lim_{x\rightarrow 0^{+}}\frac{1}{1+x}]^{2}=1$
$L=L_{1}\cdot L_{2}=1\cdot 1=1$
