수열의 극한 전체 문제

(a) To find the area $A_n$, note that the figure with $n$ sides can be divided up into $n$ isosceles triangles, each with two sides of length 1 and angle between them of $\frac{2\pi}{n}$:

Each of these triangles has area $\frac{1}{2}\sin\left(\frac{2\pi}{n}\right)$:

All together, the area of the $n$-sided figure is $A_n = \displaystyle\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$.

(b) We will discuss two ways to find $\displaystyle\lim_{n \to \infty} A_n$, which has the indeterminate form $\infty \times 0$.

First, note that as $n \to \infty$, our figures look more and more like a circle of radius 1. So, we see $A_n$ is approaching the area of a circle of radius 1. That is, $\displaystyle\lim_{n \to \infty}A_n = \pi$.

Alternately, we can make use of the limit $\lim\limits_{x \to 0}\frac{\sin x}{x}=1$. Let $x=\frac{2\pi}{n}$. Note if $n \to \infty$, then $x \to 0$.

$$\begin{aligned} \lim_{n \to \infty}A_n&=\lim_{n \to \infty}\frac{n}{2}\sin\left(\frac{2\pi}{n}\right) = \lim_{n \to \infty}\frac{\pi}{\frac{2\pi}{n}}\sin\left(\frac{2\pi}{n}\right) \\ &=\lim_{x \to 0}\pi\frac{\sin x}{x} =\pi\times 1=\pi \end{aligned}$$

(a) $f_2(x) = \begin{cases} 1 & 2 \leq x < 3\\ 0 & \text{else} \end{cases}$

(b) $f_3(x) = \begin{cases} 1 & 3 \leq x < 4\\ 0 & \text{else} \end{cases}$

(c) For any $n$, $f_n(x)=1$ for an interval of length 1, and $f_n(x)=0$ for all other $x$. So, the area under the curve is a square of side length one.

Then $A_n = \int_0^\infty f_n(x)\,\mathrm{d}x=1$ for all $n$. That is, the sequence $\{A_n\}$ is simply $\{1,1,\ldots,1\}$, a sequence of all 1s. (d) Given the description above, $\displaystyle\lim_{n \to \infty}A_n=1$. (e) For any fixed $x$, recall $\{f_n(x)\} = \{0,\ldots,0,1,0,\ldots 0,0,0,0,0,\ldots\}$. In particular, there are infinitely many zeroes at its end. So, $\displaystyle\lim_{n \to \infty} f_n(x)=0$. Then $g(x)=0$ for every $x$. (f) Given the description above, $\displaystyle \int_0^\infty g(x)\,\mathrm{d}x = \int_0^{\infty}0\,\mathrm{d}x=0$.

Remark: what we've shown here is that, for this particular $f_n(x)$, $$\lim_{n \to \infty}\int_0^{\infty}f_n(x)\,\mathrm{d}x \neq \int_0^{\infty}\lim_{n \to \infty} f_n(x)\,\mathrm{d}x$$ That is, we can't necessarily swap a limit with an integral (which is, in this case, another limit, since the integral is improper). The interested reader can look up “uniform convergence" to learn about the conditions under which these can be swapped.

The stone at position $x$ has mass $\dfrac{1}{4^x}$ kg, and we have to pull it a distance of $2^x$ metres, so the work involved in moving that one stone is $$\left(\frac{1}{4^x}\text{ kg}\right)\left(2^x\text{ m}\right)\left(9.8 ~\frac{\text{m}}{\text{sec}^2}\right) = \frac{9.8}{2^x}\text{ J}$$ Therefore, the work to move all the stones is:

$$\begin{aligned} \sum_{x=1}^{\infty}\frac{9.8}{2^x}& =\sum_{x=0}^{\infty}\frac{9.8}{2^{x+1}}\\ &=\sum_{x=0}^{\infty}\frac{9.8}{2}\cdot\frac{1}{2^{x}} =\frac{9.8}{2}\cdot\frac{1}{1-\frac12}=9.8 \text{ J} \end{aligned}$$

We consider a circle of radius $R$, with an “inner ring" from $\frac{R}{3}$ to $\frac{2R}{3}$ and an “outer ring" from $\frac{2R}{3}$ to $R$.

The area of the outer ring is: $$\pi R^2 - \pi \left(\frac{2R}{3}\right)^2=\frac{5}{9}\pi R^2$$

The area of the inner ring is: $$\pi \left(\frac{2R}{3}\right)^2 - \pi \left(\frac{R}{3}\right)^2=\frac{3}{9}\pi R^2$$

So, the ratio of the inner ring's area to the outer ring's area is $\frac{3}{5}$.

In our bullseye diagram, if we pair up any red ring with the blue ring just inside it, the blue ring has $\frac{3}{5}$ the area of the red ring. So, the blue portion of the bullseye has $\frac{3}{5}$ the area of the red portion.

Since the circle has area 1, if we let the red portion have area $A$, then $$1=A+\frac{3}{5}A=\frac{8}{5}A$$

So, the red portion has area $\frac{5}{8}$.