Find limit of a sequence

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Newcastle University Mathematics and Statistics

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

Created this.

Name	Status	Author	Last Modified
Find limit of a sequence	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Blathnaid's copy of Find limit of a sequence	draft	Blathnaid Sheridan	20/02/2017 09:42

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Question statement

Let

$x_n=\simplify[std]{({a}n+{b})/({c}n+{d})}, \quad n=1,\; 2,\; 3 \ldots$

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	2
b1	integer	3
b	integer	3
c	integer	1
d	integer	2

			Name	Type	Generated Value

n1	integer	9982
n2	integer	10009
n3	integer	9992
n4	integer	9981

			Name	Type	Generated Value

r	integer	4
ep	number	0.0001
tval	number	9998
n	integer	9998

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Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `integer`

→ Used by:

b
tval

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Parts

Part a) Gap-fill
1. Gap Mathematical expression
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Part b) Gap-fill
1. Gap Choose one from a list
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Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

What is the limit of this sequence?

$\displaystyle{\lim_{x\to\infty} x_n=\;\;}$ Unnamed gap

Input the limit as a fraction or an integer and not a decimal.

Advice

a)

The limit is $\displaystyle \simplify[std]{{a}/{c}}$ .

b)

To find the least $N$ such that all terms from the the $N$ ^th are within $10^{\var{-r}}$ of the limit, we proceed as follows:

$\begin{align} \left| \simplify[std]{x_n -({a} / {c})} \right| \leq 10^{ -\var{r}} &\iff \left| \simplify[std]{({a}n+{b})/({c}n+{d})-{a}/{c}} \right| \leq 10 ^ { -\var{r}} \\ &\iff \simplify[std]{abs({b*c-a*d})/({c^2}n+{c*d})}\leq 10 ^ { -\var{r}} \end{align}$

(We can get rid of the absolute value in the denominator as $\simplify[std]{{c^2}n+{c*d}} \gt 0$ , $\forall n=1,\; 2,\; 3 \ldots$ )

Rearranging this last inequality by multiplying both sides by $(\simplify[std]{{c^2}n+{c*d}}) \times 10^{\var{r}}$ (this is positive and so the inequality does not reverse), we get:

$\simplify[std]{{c^2}n+{c*d}} \geq \var{10^r*abs(b*c-a*d)} \iff n \geq \simplify[std]{{1}/{c^2}({10^r*abs(b*c-a*d)}-{c*d})}=\var{tval}$

{if(fract(tval)>0,"The least integer value is given by rounding up, i.e.","So")} $N=\var{N}$ .

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This question is used in the following exam:

Limits of sequences by Newcastle University Mathematics and Statistics in Content created by Newcastle University.

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Taxonomy: mathcentre

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

Parts

Gap-fill

a)

b)

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Find limit of a sequence

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

b)

Generated value: `integer`