Question:

Step 1: For $x \geq 1$, the identity $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2\tan^{-1}x$ holds. Therefore, $f(x) + 2\tan^{-1}x = \pi$. Step 2: For $x \in [0,1]$, the identities $g(x) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}x$ and $f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x$ hold. Thus, $\frac{f(x)}{g(x)} = 1$. Step 3: For $x \in (-1,0)$, $g(x) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}|x|$. Since $x \in (-1,0)$, $|x| = -x$, so $g(x) = 2\tan^{-1}(-x) = -2\tan^{-1}x$. Also, for $x \in (-1,0)$, $h(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}x$. Therefore, $g(x) + h(x) = (-2\tan^{-1}x) + (2\tan^{-1}x) = 0$. The statement "$g(x) + h(x) = \forall x \in (-1, 0)$" is syntactically incomplete and does not represent a universally true mathematical identity. Step 4: Evaluate $\lim_{x \to 1^+}(f(x)+g(x)+h(x))$. For $x > 1$: $f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2\tan^{-1}x$. $$ \lim_{x \to 1^+} f(x) = \pi - 2\tan^{-1}(1) = \pi - 2\left(\frac{\pi}{4}\right) = \pi - \frac{\pi}{2} = \frac{\pi}{2} $$ $g(x) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}x$. $$ \lim_{x \to 1^+} g(x) = 2\tan^{-1}(1) = 2\left(\frac{\pi}{4}\right) = \frac{\pi}{2} $$ $h(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$. For $x > 1$, $h(x) = 2\tan^{-1}x - \pi$. $$ \lim_{x \to 1^+} h(x) = 2\tan^{-1}(1) - \pi = 2\left(\frac{\pi}{4}\right) - \pi = \frac{\pi}{2} - \pi = -\frac{\pi}{2} $$ Thus, $\lim_{x \to 1^+}(f(x)+g(x)+h(x)) = \frac{\pi}{2} + \frac{\pi}{2} - \frac{\pi}{2} = \frac{\pi}{2}$. Step 5: Evaluate $\lim_{x \to 1^-}(f(x)+g(x)+h(x))$. For $x < 1$ (and $x>0$): $f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x$. $$ \lim_{x \to 1^-} f(x) = 2\tan^{-1}(1) = 2\left(\frac{\pi}{4}\right) = \frac{\pi}{2} $$ $g(x) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}x$. $$ \lim_{x \to 1^-} g(x) = 2\tan^{-1}(1) = 2\left(\frac{\pi}{4}\right) = \frac{\pi}{2} $$ $h(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}x$. $$ \lim_{x \to 1^-} h(x) = 2\tan^{-1}(1) = 2\left(\frac{\pi}{4}\right) = \frac{\pi}{2} $$ Thus, $\lim_{x \to 1^-}(f(x)+g(x)+h(x)) = \frac{\pi}{2} + \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{2}$. Step 6: Compute the ratio of the limits. $$ \frac{\lim_{x \to 1^+}(f(x)+g(x)+h(x))}{\lim_{x \to 1^-}(f(x)+g(x)+h(x))} = \frac{\frac{\pi}{2}}{\frac{3\pi}{2}} = \frac{1}{3} $$