수열의 극한 전체 문제

${a}_{n}$ is decreasing and bounded below by $2.$ The limit $a$ must satisfy $a=\sqrt{2a}$ so $a=2,$ independent of the initial value.

$\sum_{n=1}^{\infty }{a}_{n}=f(1)-f(2)$

$5/9$

The part of the first dose after $n$ hours is $d{r}^{n},$ the part of the second dose is $d{r}^{n-N},$ and, in general, the part remaining of the $m\text{th}$ dose is $d{r}^{n-mN},$ so

$A(n)=\sum_{l=0}^{m}d{r}^{n-lN}=\sum_{l=0}^{m}d{r}^{k+(m-l)N}=\sum_{q=0}^{m}d{r}^{k+qN}=d{r}^{k}\sum_{q=0}^{m}{r}^{Nq}=d{r}^{k}\frac{1-{r}^{(m+1)N}}{1-{r}^{N}},n=k+mN.$

Let's take stock: $\sin(1/n) \to \sin (0)=0$ as $n \to \infty$, so $\log\left(\sin(1/n)\right) \to -\infty$. However, $\log(2n) \to \infty$. So, we have some tension here: the two pieces behave in ways that pull the terms of the sequence in different directions. (Recall we cannot conclude anything like “$-\infty+\infty=0$.")

We try using logarithm rules to get a clearer picture.

$$\begin{aligned} \log\left(\sin\frac{1}{n}\right)+\log(2n)&=\log\left(2n\sin\left(\frac1n\right)\right) \end{aligned}$$

Still, we have indeterminate behaviour: $2n\sin(1/n)$ is the product of $2n$, which grows without bound, and $\sin(1/n)$, which approaches zero. In the past, we learned that we can handle the indeterminate form $0\cdot\infty$ with l'Hopital's rule (after a little algebra), but there's a slicker way. Note $1/n \to 0$ as $n \to \infty$. If we write $\frac{1}{n}=x$, then this piece of our limit resembles something familiar.

$$\begin{aligned} 2n\sin\left(\frac{1}{n}\right)&=2\left(\frac{\sin x}{x}\right) \\ \text{If $n \to \infty$, then $x = \frac{1}{n}\to 0$.} \\ \lim_{n \to \infty}2n\sin\left(\frac{1}{n}\right)&=2\lim_{x \to 0}\frac{\sin x}{x} \\ \text{That limit is familiar:} \\ &=2(1)=2 \\ \text{Then:} \\ \lim_{n \to \infty}\log\left(2n\sin\left(\frac{1}{n}\right)\right)&=\log 2 \end{aligned}$$

Note: if you have forgotten that $\displaystyle\lim_{x \to 0}\frac{\sin x}{x}=1$, you can also evaluate this limit using l'Hopital's rule:

$$\begin{aligned} \underbrace{\lim_{x \to 0}\frac{\sin x}{x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to 0}{\mathrm{den}\to0}} &=\lim_{x \to 0}\frac{\cos x}{1}=\cos 0 =1 \end{aligned}$$

First, although this sequence is not defined for some small values of $n$, it is defined as long as $n \geq 5$, so it's not a problem to take the limit as $n \to \infty$. Second, we notice that our limit has the indeterminate form $\infty-\infty$. Since this form is indeterminate, more work is needed to find our limit, if it exists.

A standard trick we saw last semester with functions of this form was to multiply and divide by the conjugate of the expression, $\sqrt{n^2+5n}+\sqrt{n^2-5n}$. Then the denominator will be the sum of two similar things, rather than their difference. See the work below to find out why that is helpful.

$$\begin{aligned} \sqrt{n^2+5n}-\sqrt{n^2-5n}&=\big(\sqrt{n^2+5n}-\sqrt{n^2-5n}\big)\left(\frac{\sqrt{n^2+5n}+\sqrt{n^2-5n}}{\sqrt{n^2+5n}+\sqrt{n^2-5n}}\right)\\ &=\frac{(n^2+5n)-(n^2-5n)}{\sqrt{n^2+5n}+\sqrt{n^2-5n}}\\ &=\frac{10n}{\sqrt{n^2+5n}+\sqrt{n^2-5n}} \\ \text{Now, we'll cancel out $n$ from the top and the bottom. Note $n=\sqrt{n^2}$.} \\ &=\frac{10n}{\sqrt{n^2+5n}+\sqrt{n^2-5n}}\left(\frac{\frac1n}{\frac1n}\right)\\ &=\frac{10n}{\sqrt{n^2+5n}+\sqrt{n^2-5n}}\left(\frac{\frac1n}{\frac{1}{\sqrt{n^2}}}\right)\\ &=\frac{10}{\sqrt{1+\frac5n}+\sqrt{1-\frac5n}} \\ \text{Now, the limit is clear.} \\ \lim_{n \to \infty}\frac{10}{\sqrt{1+\frac5n}+\sqrt{1-\frac5n}}&=\frac{10}{\sqrt{1+0}+\sqrt{1+0}}=\frac{10}{1+1}=5 \end{aligned}$$

First, we note that we have in indeterminate form: as $n$ grows, $2+\frac1n \to 2$, so $n\left[\left(2+\frac1n\right)^{100}-2^{100}\right]$ has the form $\infty \cdot 0$. To overcome this difficulty, we could use some algebra and l'Hopital's rule, but there's a slicker way. If we let $h = \frac{1}{n}$, then $h \to 0$ as $n\to\infty$, and our limit looks like:

$$\begin{aligned} \lim_{n \to \infty}n\left[\left(2+\frac1n\right)^{100}-2^{100}\right]&= \lim_{h \to 0}\frac{\left(2+h\right)^{100}-2^{100}}{h} \\ \text{This reminds us of the definition of a derivative.} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{x^{100}\right\}&=\lim_{h \to 0}\frac{(x+h)^{100}-x^{100}}{h} \end{aligned}$$

So, if we set $f(x)=x^{100}$, our limit is simply $f'(2)$. That is, $\left[100x^{99}\right]_{x=2}= 100\cdot 2^{99}$.

(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.