극한·미분 심화 전체 문제

(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 $$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} x^{2x} = 1$$

(a) We note that both $\ln(x)$ or $\sqrt{x}$ increase without bound as $x \to \infty$, so we consider $$\lim_{x \to \infty}\frac{\ln(x)}{\sqrt{x}}\,.$$ By L'Hôpital's Rule and some simplifying algebra, $$\lim_{x \to \infty}\frac{\ln(x)}{\sqrt{x}} = \lim_{x \to \infty}\frac{\frac{1}{x}}{\frac{1}{2x^{1/2}}} = \lim_{x \to \infty}\frac{2x^{1/2}}{x} = \lim_{x \to \infty}\frac{2}{x^{1/2}}\,.$$ Thus, it follows that $$\lim_{x \to \infty}\frac{\ln(x)}{\sqrt{x}} = 0\,,$$ and hence $\sqrt{x}$ dominates $\ln(x)$. (b) Using a similar argument to our work in $(a)$, we can show that $\sqrt[n]{x}$ dominates $\ln(x)$ for any positive integer $n$. In particular, by L'Hôpital's Rule and some simplifying algebra, $$\lim_{x \to \infty}\frac{\ln(x)}{\sqrt[n]{x}} = \lim_{x \to \infty}\frac{\frac{1}{x}}{\frac{1}{nx^{(n-1)/n}}} = \lim_{x \to \infty}\frac{nx^{(n-1)/n}}{x}\,.$$ Since $\frac{x^{(n-1)/n}}{x} = x^{(n-1)/n - 1} = x^{(n-1-n)/n} = x^{-1/n}$, it follows that $$\lim_{x \to \infty}\frac{\ln(x)}{\sqrt[n]{x}} = \lim_{x \to \infty} \frac{1}{x^{1/n}} = 0$$ we've shown that $\sqrt[n]{x}$ dominates $\ln(x)$ for any positive integer $n$. (c) To see why $e^x$ will dominate any polynomial function, consider $p(x) = x^4$. If we consider the limit of their quotient and apply LHR once, we see that $$\lim_{x \to \infty} \frac{x^4}{e^x} = \lim_{x \to \infty} \frac{4x^3}{e^x}$$ By repeated application of LHR, the numerator (regardless of what polynomial we start with) will eventually be simply a constant (after $n$ applications of LHR), and thus with $e^x$ still in the denominator, the overall limit will be $0$. (d) Here we naturally consider the limit of the quotient of $\ln(x)$ and $x^n$, where $n \ge 1$. Since both functions increase without bound, we can apply LHR. Doing so along with some simplifying algebra, we see $$\lim_{x \to \infty} \frac{\ln(x)}{x^n} = \lim_{x \to \infty} \frac{\frac{1}{x}}{nx^{n-1}} = \frac{1}{nx^n} = 0$$ and thus $x^n$ indeed dominates. (e) Consider $f(x) = 3x^2 + 1$ and $g(x) = -0.5x^2 + 5x - 2$. Since these functions are polynomials of the same degree, we know $$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{3}{-0.5} = -6\,.$$