1. The problem involves finding the second derivative $h''(x)$ of the function $h(x) = \sqrt[n]{nx} = (nx)^{\frac{1}{n}}$ and analyzing its behavior as $n \to \infty$. 2. Recall the power rule for derivatives: if $f(x) = x^m$, then $f'(x) = m x^{m-1}$. 3. First, rewrite $h(x)$ as $h(x) = n^{\frac{1}{n}} x^{\frac{1}{n}}$. 4. Differentiate once: $$h'(x) = n^{\frac{1}{n}} \cdot \frac{1}{n} x^{\frac{1}{n} - 1} = n^{\frac{1}{n} - 1} x^{\frac{1}{n} - 1}$$ 5. Differentiate a second time: $$h''(x) = n^{\frac{1}{n} - 1} \cdot \left(\frac{1}{n} - 1\right) x^{\frac{1}{n} - 2} = n^{\frac{1}{n} - 1} \left(\frac{1}{n} - 1\right) x^{\frac{1}{n} - 2}$$ 6. Simplify the expression: $$h''(x) = n^{\frac{1}{n} - 1} \left(\frac{1 - n}{n}\right) x^{\frac{1}{n} - 2} = (1 - n) n^{\frac{1}{n} - 2} x^{\frac{1}{n} - 2}$$ 7. Analyze the limit as $n \to \infty$: - $n^{\frac{1}{n}} \to 1$ because $\lim_{n \to \infty} n^{1/n} = 1$. - So $n^{\frac{1}{n} - 2} = n^{\frac{1}{n}} \cdot n^{-2} \to 1 \cdot 0 = 0$. - The factor $(1 - n)$ tends to $-\infty$. - The term $x^{\frac{1}{n} - 2} \to x^{-2}$. 8. Combining these, the product tends to: $$h''(x) \approx (1 - n) \cdot 0 \cdot x^{-2} = 0$$ because the $n^{-2}$ term dominates the linear growth of $(1-n)$. 9. Therefore, as $n \to \infty$, $h''(x) \to 0$. **Final answer:** $$\lim_{n \to \infty} h''(x) = 0$$