Find limits of sequences

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Newcastle University Mathematics and Statistics

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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Name	Status	Author	Last Modified
Find limits of sequences	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Ben's copy of Find limits of sequences	draft	Ben Brawn	13/10/2017 02:43

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Question statement

What are the following limits?

Give any introductory information the student needs.

			Name	Type	Generated Value

r	integer	8
v15	decimal	dec("0.56234")
v110	decimal	dec("0.34485")
v150	decimal	dec("0.14542")

			Name	Type	Generated Value

k1	integer	158842
v25	decimal	dec("1.12722")
v210	decimal	dec("1.00240")
v250	decimal	dec("1.00000")

			Name	Type	Generated Value

k	number	8
v35	number	1.02101
v310	number	1.00042
v350	number	1.00000

			Name	Type	Generated Value

c	integer	-14
d	integer	5
al	integer	2
ga	integer	4
v45	decimal	dec("-7.11735")
v410	decimal	dec("-7.00230")
v450	decimal	dec("-7.00000")

			Name	Type	Generated Value

n	number	18
v55	number	0.08101
v510	number	-0.00068
v550	number	0.00000
v560	number	0.00000

			Name	Type	Generated Value

a3	integer	2
ab	number	2
b3	integer	7
v65	number	1.32539
v610	number	1.33017
v650	number	1.33066
v660	number	1.33071
val	decimal	dec("1.3307")
tol	integer	0
valexp	decimal	dec("1.33071")

			Name	Type	Generated Value

			a	integer	3
			b	integer	13

			Name	Type	Generated Value

b1	integer	5
be	integer	-5
c1	integer	-8
d1	integer	2
d2	integer	2
de	integer	5
s1	integer	-1
s5	integer	1

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(2..10)

JME function reference

Description

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

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

v110
v15
v150

This variable doesn't seem to be used anywhere.

Parts

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Gap-fill

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

$\displaystyle{\lim_{n \to \infty}\left(\frac{1}{n^{1/\var{r}}}\right)=}$ Unnamed gap

Advice

All calculations below are rounded to $5$ decimal places.

The notation $a \approx b$ means that $a$ and $b$ are approximately equal.

a)

Using a calculator for $3$ values of $n$ :

$n$	$\displaystyle{\frac{1}{n^{1/\var{r}}}}$
$100$	$\var{v15}$
$5000$	$\var{v110}$
$5000000$	$\var{v150}$

This indicates that $\displaystyle{\lim_{n \to \infty}\left(\frac{1}{n^{1/\var{r}}}\right)=0}$

In fact $\displaystyle{\lim_{n \to \infty}\left(\frac{1}{n^r}\right)=0}$ for any $r \gt 0$

b)

$n$	$\displaystyle{\var{k1}^{1/n}}$
$100$	$\var{v25}$
$5000$	$\var{v210}$
$5000000$	$\var{v250}$

This indicates that $\displaystyle{\lim_{n \to \infty}\var{k1}^{1/n}=1}$ .

c)

$n$	$\displaystyle{\var{k}^{1/n}}$
$100$	$\var{v35}$
$5000$	$\var{v310}$
$5000000$	$\var{v350}$

This indicates that $\displaystyle{\lim_{n \to \infty}\var{k}^{1/n}=1}$ .

From the last two questions it seems that $\displaystyle{\lim_{n \to \infty} k^{1/n}=1}$ for any $k \gt 0$ – and this is in fact true.

d)

$n$	$\displaystyle{\frac{\var{c}n+\var{d}}{\var{al}n-\var{ga}}}$
$100$	$\var{v45}$
$5000$	$\var{v410}$
$5000000$	$\var{v450}$

This indicates that $\displaystyle \lim_{n \to \infty}\left(\simplify[std]{({c}n+{d})/({al}n-{ga})}\right) = \simplify[std]{{c}/{al}}$ .

In general, $\displaystyle{ \lim_{n \to \infty}\left(\frac{an+b}{cn+d}\right)= \frac{a}{c} }$ when $c \neq 0$ .

e)

$n$	$\displaystyle{\left(\simplify{{c}/{n}}\right)^n}$
$10$	$\var{v55}$
$29$	$\var{v510}$
$50$	$\var{v550}$
$89$	$\var{v560}$

This indicates that $\displaystyle{\lim_{n \to \infty}\left(\simplify{{c}/{n}}\right)^n}= 0$ .

In general $\displaystyle{\lim_{n \to \infty} r^n= 0}$ if $|r| \lt 1$ .

f)

We have the limit:

$\lim_{n\to\infty}\left(1+\frac{a}{n}\right) = e^a$

The following table confirms that the values are converging to $\displaystyle{\simplify[std]{e^({a3}/{b3})={valexp}}}$ (to five decimal places).

$n$	$\displaystyle{\left(\simplify[std]{1+{a3}/({b3}n)}\right)^n}$
$10$	$\var{v65}$
$100$	$\var{v610}$
$1000$	$\var{v650}$
$10000$	$\var{v660}$

Hence the answer asked for is $\var{val}$ to $4$ decimal places.

g)

The answer to this question is based upon neglecting terms in polynomials in $n$ for large $n$ .

For example, $n^3+1000000n^2+1000000000 \approx n^3$ for large $n$ because the $n^3$ term completely dominates the other terms as $n \to \infty$ .

A more precise way of saying this is:

$\lim_{n\to\infty}\left(\frac{n^3+1000000n^2+1000000000}{n^3}\right)=1$

So for large $n$ ,

$\begin{align} \frac{\left(\simplify[std]{{al^d}n^({a*d})+{be}n^{b}+{c}}\right)^{1/\var{d}}} {\left(\simplify[std]{{ga^d1}n^({a*d1})+{de}n^{b1}+{c1}}\right)^{1/\var{d1}}} &\approx \frac{\left(\simplify[std]{{al^d}n^({a*d})}\right)^{1/\var{d}}} {\left(\simplify[std]{{ga^d1}n^({a*d1})}\right)^{1/\var{d1}}} \\ \\ &= \frac{\simplify[std]{{al^d}^(1/{d})n^{a}}} {\simplify[std]{{ga^d1}^(1/{d1})n^{a}}} \\ \\ &= \simplify[std]{{al}/{ga}} \end{align}$

Hence the limit is $\displaystyle{\simplify[std]{{al}/{ga}}}$

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

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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)

c)

d)

e)

f)

g)

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Find limits of sequences

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)

c)

d)

e)

f)

g)

Generated value: `integer`