November 22, 2021 – Physicspupil

November 22, 2021October 24, 2022

202111221624 Exercises 9.1.A (Q2)

For Exercises 1-8, determine if the given sequence is convergent. If so then find its limit.

2. $\displaystyle{\bigg\{ \frac{n^2}{3n^2+7n-2}\bigg\}_{n=1}^{\infty} }$

Solution.

By L’Hôpital’s Rule, treating an integer $n \ge 1$ as a real-valued variable $x$ ,

$\begin{aligned} &\quad \lim_{n\to\infty}\frac{n^2}{3n^2+7n-2} \\ & = \lim_{n\to\infty}\frac{2n}{6n+7} \\ & = \frac{2}{6} \\ & = \frac{1}{3}\\ \end{aligned}$

Thus the sequence is convergent and its limit is $1/3$ .

November 22, 2021October 24, 2022

202111221615 Exercises 9.1.A (Q1)

For Exercises 1-8, determine if the given sequence is convergent. If so then find its limit.

$\{ ne^{-n}\}_{n=0}^{\infty}$

Solution.

By L’Hôpital’s Rule, treating an integer $n \ge 0$ as a real-valued variable $x$ ,

$\begin{aligned} & \quad \lim_{n\to\infty} \frac{n}{e^n} \\ & = \lim_{n\to\infty} \frac{0}{e^n} \\ & = 0 \\ \end{aligned}$

Thus the sequence is convergent and its limit is $0$ .

November 22, 2021October 24, 2022

202111221509 Solution to 1971-CE-AMATH-1

When the functions $x^2-bx+3$ and $2b-x$ are divided by $(x-a)$ , the remainders are 1 and 4 respectively. Find $a$ and $b$ .

Solution.

Translating mathematically, we have

$\begin{aligned} x^2-bx+3 & = (x-a)F(x)+1 \\ 2b-x & = (x-a)G(x)+4\\ \end{aligned}$

Plugging in $x=a$ , we have

$\begin{aligned} (a)^2-b(a)+3 & = \big( (a)-a\big) F(a)+1 \\ 2b-(a) & = \big( (a)-a\big) G(a)+4 \\ \end{aligned}$

After simplifying it, we have

$\begin{aligned} a^2-ab & = -2\\ -a+2b & = 4\\ \end{aligned}$

Solving two equations with two unknowns,

$\begin{cases} & a =2\\ & b =3 \\ \end{cases}$

is the solution.

M	T	W	T	F	S	S
1	2	3	4	5	6	7
8	9	10	11	12	13	14
15	16	17	18	19	20	21
22	23	24	25	26	27	28
29	30