극한·미분 심화 전체 문제

(a) Since $e$ raised to any power is always positive (and thus never $0$), it follows that the only value that makes $f'(x) = (x-1)e^{-x^2} = 0$ is $x=1$. This is confirmed by plotting $f'$ where we only see the graph cross the $x$-axis at $x=1$ (though we do see the graph approach the $x$-axis) as $x$ increases or decreases without bound). (b) We know that $f''(x) > 0$ wherever $f'$ is increasing. From the graph of $f'$, we see that $f'$ is increasing approximately on the interval $-0.37 < x < 1.37$, and thus $f$ is concave up on this interval. (c) The tangent line approximation at $a = 2$ is $L(x) = f(2) + f'(2)(x-2) \approx -3 + 0.01832(x-2)$. Thus, $f(1.88) \approx L(1.88) \approx -3 + 0.01832(1.88-2) = -3.00022$. Since $f''(x) < 0$ at and around $x = 2$, $f$ is concave down there, and therefore the tangent line lies above the curve. This makes our estimate larger than the true value of $f(1.88)$.

See the figure below for possible graphs of $f$ and $g$. To evaluate the limit, we can apply L'Hôpital's Rule twice. Since $f(3) = g(3) = 0$, we know by the rule that

$$\lim_{x \to 3} \frac{f(x)}{g(x)} = \lim_{x \to 3} \frac{f'(x)}{g'(x)}\,.$$ Since $f'(3) = g'(3) = 0$, the limit is still indeterminate, and we can apply the rule again to get

$$\lim_{x \to 3} \frac{f(x)}{g(x)} = \lim_{x \to 3} \frac{f''(x)}{g''(x)} = \frac{f''(3)}{g''(3)} = \frac{-2}{1}\,.$$ Hence, $\lim_{x \to 3} h(x) = -2$.

First, we note that since both the numerator and denominator of $R$ are quadratic, the limit as $x \to \infty$ can be found by taking the quotient of the leading coefficients. In particular,

$$\lim_{x \to \infty} R(x) = \frac{3}{5}$$

and thus the horizontal asymptote of $R$ is $y = \frac{3}{5}$.

Next, since $b$ and $c$ are distinct, we see that the numerator of $R$ is zero (and the denominator is not) when $x = b$, and thus $R(b) = 0$. In addition, the denominator is zero (and the numerator is not) when $x = c$, and this results in there being a vertical asymptote at $x = c$.

Finally, we observe that $x = a$ makes both the numerator and denominator zero. Since $R(a)$ is thus not defined, the graph of $R$ has either a hole or a vertical asymptote at $x = a$. To determine which, we evaluate the limit and see that

$$\lim_{x \to a} \frac{3(x-a)(x-b)}{5(x-a)(x-c)} = \lim_{x \to a} \frac{3(x-b)}{5(x-c)} = \frac{3(a-b)}{5(a-c)}\,.$$ This tells us that $R$ has a hole at the point $(a, \frac{3(a-b)}{5(a-c)})$. Moreover, $R$ is not continuous at $x = a$ and $x = c$.

A possible graph of $R$ is shown in the following figure.

(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 $$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{3}{-0.5} = -6\,.$$