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(a) Since $h(x) = \ln(g(x))$ and we are given that $g(x) = x^{2x}$, it follows that $$h(x) = \ln(x^{2x}) = 2x \cdot \ln(x)\,.$$ (b) Since $x = (x^{-1})^{-1} = \frac{1}{\frac{1}{x}}$, we can also write $$h(x) = \frac{2\ln(x)}{\frac{1}{x}}\,.$$ (c) Since $\ln(x) \to -\infty$ as $x \to 0^+$ and $\frac{1}{x} \to +\infty$ as $x \to 0^+$, we see that $$\lim_{x \to 0^+} h(x) = \lim_{x \to 0^+} \frac{2\ln(x)}{\frac{1}{x}}$$ is of an indeterminate form to which we can apply L'Hôpital's Rule. Doing so, $$\lim_{x \to 0^+} h(x) = \lim_{x \to 0^+} \frac{2\ln(x)}{\frac{1}{x}} = \lim_{x \to 0^+} \frac{\frac{2}{x}}{-\frac{1}{x^2}}$$ Simplifying algebraically, $$\lim_{x \to 0^+} h(x) = \lim_{x \to 0^+} \frac{2}{x} \cdot \left(-\frac{x^2}{1}\right) = \lim_{x \to 0^+} (-2x)$$ Hence, it follows that $\lim_{x \to 0^+} h(x) = 0$. (d) Having shown that $\lim_{x \to 0^+} h(x) = 0$ and recalling that $h(x) = \ln(g(x))$, we know that as $x \to 0^+$, the natural log of $g(x)$ tends to $0$. Since $\ln(1) = 0$, it must be the case that $g(x)$ is tending to $1$, and thus $$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} x^{2x} = 1$$