1. **State the problem:** We need to find the limit $$\lim_{x \to 1^+} (x^2 - 1) \tan \frac{\pi x}{2}$$. 2. **Recall the formula and behavior:** As $x \to 1^+$, note that $x^2 - 1 = (x-1)(x+1)$ approaches 0 since $x-1 \to 0^+$. The tangent function $\tan \frac{\pi x}{2}$ approaches $+\infty$ or $-\infty$ depending on the side of approach because $\frac{\pi x}{2} \to \frac{\pi}{2}^+$, where tangent has a vertical asymptote. 3. **Rewrite the expression:** $$ (x^2 - 1) \tan \frac{\pi x}{2} = (x-1)(x+1) \tan \frac{\pi x}{2} $$ 4. **Substitute $x = 1 + h$ with $h \to 0^+$:** $$ (1+h - 1)(1+h + 1) \tan \frac{\pi (1+h)}{2} = h (2 + h) \tan \left( \frac{\pi}{2} + \frac{\pi h}{2} \right) $$ 5. **Use the tangent shift identity:** $$ \tan \left( \frac{\pi}{2} + \theta \right) = -\cot \theta $$ So, $$ \tan \left( \frac{\pi}{2} + \frac{\pi h}{2} \right) = -\cot \left( \frac{\pi h}{2} \right) $$ 6. **Rewrite the limit:** $$ \lim_{h \to 0^+} h (2 + h) \left(-\cot \frac{\pi h}{2} \right) = - \lim_{h \to 0^+} h (2 + h) \cot \frac{\pi h}{2} $$ 7. **Approximate $\cot$ near 0:** $$ \cot z \approx \frac{1}{z} \quad \text{as} \quad z \to 0 $$ So, $$ \cot \frac{\pi h}{2} \approx \frac{2}{\pi h} $$ 8. **Substitute approximation:** $$ - \lim_{h \to 0^+} h (2 + h) \cdot \frac{2}{\pi h} = - \lim_{h \to 0^+} (2 + h) \frac{2}{\pi} = - \frac{4}{\pi} $$ 9. **Final answer:** $$ \boxed{- \frac{4}{\pi}} $$ This means the limit exists and equals $-\frac{4}{\pi}$ as $x$ approaches 1 from the right.