Sequences and Limits 1.1

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Name	Status	Author	Last Modified
Sequences and Limits 1.1	Ready to use	Bill Foster	01/06/2016 09:47
Katie's copy of Sequences and Limits 1.1	Ready to use	Katie Dexter	01/06/2016 09:47
Sequences and Limits 1.1	draft	Marlon Arcila	31/08/2016 17:31

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Question statement

What are the following limits?

Give any introductory information the student needs.

			Name	Type	Generated Value

ab	number	3
valexp	decimal	dec("1.45499")
v35	number	1.02558
al	integer	4
v15	decimal	dec("0.21544")
v150	decimal	dec("0.00585")
v610	number	1.45397
b1	integer	11
ga	integer	4
b3	integer	8
d2	integer	4
d1	integer	4
val	decimal	dec("1.4550")
s1	integer	-1
v110	decimal	dec("0.05848")
v560	number	-0.00021
s5	integer	1
v650	number	1.45489
k1	integer	160041
r	integer	3
v45	decimal	dec("-10.09596")
tol	integer	0
v210	decimal	dec("1.00240")
v250	decimal	dec("1.00000")
be	integer	-3
v25	decimal	dec("1.12731")
v65	number	1.44504
de	integer	-1
v450	decimal	dec("-10.00000")
a3	integer	3
c1	integer	-7
d	integer	2
a	integer	3
c	integer	-40
b	integer	3
v550	number	0.00852
v350	number	1.00000
v510	number	-0.06304
k	number	12.5
v410	decimal	dec("-10.00190")
n	number	44
v55	number	0.38554
v310	number	1.00051
v660	number	1.45498

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Name

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Value

xxxxxxxxxx
 
abs(a3)

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Generated value: `number`

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Parts

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Part b) Gap-fill
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Part c) Gap-fill
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Part d) Gap-fill
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Part f) Gap-fill
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Gap-fill

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

Advice

All calculations below are 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 $\displaystyle{\lim_{n \to \infty}\var{k1}^{1/n}=1}$ , see next question as well.

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 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

$\lim_{n \to \infty}\left(\frac{an+b}{cn+d}\right)= \frac{a}{c}$ 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 (five decimal places)

$\displaystyle{\simplify[std]{e^({a3}/{b3})={valexp}}}$

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

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 $n$ as the $n^3$ term completely dominates the other terms as $n \longrightarrow \infty$ .

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 $n$

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

$\displaystyle{\simplify[std]{{al}/{ga}}}$

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Christian Lawson-Perfect 10 years, 2 months ago

Bill Foster 55 years, 5 months ago

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b)

c)

d)

e)

f)

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Sequences and Limits 1.1

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 10 years, 2 months ago

Bill Foster 55 years, 5 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

a)

b)

c)

d)

e)

f)

g)

Generated value: `number`