Question

solve the inequality |(n)/(n+1)-1|<0.01

Answer 1

Hint :

\[\frac{n}{n+1}-1=\frac{n}{n+1}-\frac{n+1}{n+1}=-\frac{1}{n+1}\]

Answer 2

abs cancels the negative

Answer 3

i'm stuck with solving for <0.01

Answer 4

better to reverse it from first step\[|\frac{1}{n+1}|<\frac{1}{100}\]\[|n+1|>100\]now solve for n

Answer 5

i can see how to come up with the answer of n = 99 but i'm trying to work out the reasoning for the answer given at the value of 101

Answer 6

that answer given does include 99

Answer 7

conider the equation\[|x|>a\]where \(a\) is a positive real number...

solution of this inequality is \(x>a\) and \(x<-a\)

Answer 8

forgive me if my question is stupid, in being absolute value i am aware of the rule that you have just stated, but is there working to show

Answer 9

i should also state that the answer is given as n>99 or n<-101, i think i just answered my own question

Answer 10

yep

Answer 11

do u wanna know how can we prove that rule?

Answer 12

yes if you don't mind

Answer 13

well consider 2 cases if \(x>0\) then \(|x|=x\) and \(x>a\)

if \(x<0\) then \(|x|=-x\) and \(-x>a\) multiply by \(-1\) ----> \(x<-a\)

Answer 14

i'd say that's pretty much the proven

Answer 15

you got a new fan

Answer 16

lol...thank u

Answer 17

and a similar rule : when \(|x|<a\) where \(a\) is a positive real number...

then u can prove : \(-a<x<a\)

Answer 18

ok hang on i gotta take all this down and go over it a few times to make it sink in, i know these rules will be handy for the future, i got a lot of catching up to do here in calculus

Answer 19

i still can't find how to show the value of 101

Answer 20

oh sorry i was out

Answer 21

i mean -101

Answer 22

we got \(|n+1|>100\) right?

Answer 23

yea

Answer 24

well according to the first rule ...

\(n+1>100\) or \(n+1<-100\) from this

n+1>100 ----> n>99

n+1<-100 -----> n<-101

Answer 25

now i can see where the rule fits,

Answer 26

:)

Answer 27

thanks again, you'eve really taken the timer to explain things, i got another one that i have been working on its a bit harder than that one should i post it or can i just give it to you here

Answer 28

np...give it here

Answer 29

|(2n^2)/(n^2+4)(-2)|

Answer 30

\[|\frac{2n^2}{(n^2+4)(-2)}|\]?

Answer 31

the -2 is separate its not part of the denominator

Answer 32

\[\left| \frac{2n^2}{(n^2+4)}(-2) \right|\]

Answer 33

thats better

Answer 34

u can get constants out of abs whatever it is so i can write\[\left| \frac{2n^2}{(n^2+4)}(-2) \right|=4\left| \frac{n^2}{(n^2+4)} \right|\]

Answer 35

in better words : \[\left| \frac{2n^2}{(n^2+4)}(-2) \right|=|-4|\left| \frac{n^2}{(n^2+4)} \right|=4\left| \frac{n^2}{(n^2+4)} \right|\]

Answer 36

because we know that \(|ab|=|a||b|\)

Answer 37

i can honestly say that i have no idea where this is leading

Answer 38

i'll tell u what they mean by this question... is it ok till there?

Answer 39

yes but because i've never done this before i'm frozen here, i'm entirely dependant on your lead here

Answer 40

:)

Answer 41

im going to prove\[\frac{n^2}{n^2+4}>0\]for all real numbers \(n\)

Answer 42

can u tell me why?

Answer 43

because x>a

Answer 44

where a is a positive real number

Answer 45

then |x|=x, yes , no|?|

Answer 46

yep

Answer 47

if we prove that then we can omit the abs sign

Answer 48

i thought i could understand at least that much

Answer 49

lets go to work on that \[\frac{n^2}{n^2+4}>0\]

Answer 50

im trying but my paper is getting thin from rubbing out the mistakes

Answer 51

\(n^2\) and \(n^2+4\) are positive values... so that fraction will be positive also

Answer 52

i agree

Answer 53

so the final result will be\[\left| \frac{2n^2}{(n^2+4)}(-2) \right|=|-4|\left| \frac{n^2}{(n^2+4)} \right|=4\left| \frac{n^2}{(n^2+4)} \right|=\frac{4n^2}{n^2+4}\]

Answer 54

Mmm this is a hard one

Answer 55

but u have it now ;)

Answer 56

i,m still trying to find my way to the answer given at n>sqrt396 or n<-sqrt396

Answer 57

because i don,t know how to deal with the quadratic

Answer 58

so what is the original question?

Answer 59

i think u missed something in the question

Answer 60

solve the following inequalities

Answer 61

is all that is missing

Answer 62

u gave me \[\left| \frac{2n^2}{(n^2+4)}(-2) \right|\] well greater or less than what?

Answer 63

oops <0.01

Answer 64

so now the problem is\[\ \frac{4n^2}{(n^2+4)}<0.01\]

Answer 65

sorry i been doing math since 8 o clock this morning

Answer 66

yes

Answer 67

no problem :)

Answer 68

reverse both sides of it what will u get?

Answer 69

i would only be guessing ,

Answer 70

i hate to butt in, and i hope i am wrong, but i am going to bet that it was
\[\left| \frac{2n^2}{(n^2+4)}-2 \right|\] and not
\[\left| \frac{2n^2}{(n^2+4)}(-2) \right|\]

Answer 71

like i said, i really hope i am wrong, because this is such good work

Answer 72

i am thinking that you are trying to prove, using the definition, that the limit as x goes to zero is 2
that would give the inequality i wrote above

Answer 73

u r right ...thats it :)

Answer 74

then it is not a multiplication, but a subtraction, and the algebra will be different

Answer 75

you are going to get
\[|\frac{-8}{n^2+4}|<0.1\] which is much easier

Answer 76

i would really like to see how you found -8

Answer 77

algebra

Answer 78

\[\frac{2n^2}{n^2+4}-2=\frac{2n^2-2(n^2+4)}{n^2+4}\]

Answer 79

when you wrote
\[\left| \frac{2n^2}{(n^2+4)}(-2) \right|\] @mukushla understandably thought it was a product because of the parentheses around the \(-2\) but i think in fact it should have been
\[\left| \frac{2n^2}{(n^2+4)}-2 \right|\] which is entirely different

Answer 80

i have obviously been to much math today, satellite looks like your running with it

Answer 81

wait that is wrong scratch that

Answer 82

you are trying to prove
\[\lim_{n\to \infty}\frac{2n^2}{n^2+4}=2\]

Answer 83

no my limit questions are coming up next, we are just trying to solve the inequality have you got any more input

Answer 84

you are being gently led it to that topic. this is the algebra you will have to do when you get there

Answer 85

great lead on

Answer 86

\[|\frac{-8}{n^2+4}|<0.1\]
\[|\frac{8}{n^2+4}|<0.1\]
\[-0.1<\frac{8}{n^2+4}<.1\]
gives two inequalities to solve, just like the first one
\[-.1<\frac{8}{n^2+4}\] and
\[\frac{8}{n^2+4}<.1\]
normally you cannot multiply with a variable, because you do not know if it is positive or negative
but in this case since \(n^2+4>0\) you can solve by multiplying both sides by \(n^2+4\)

Answer 87

do you think we can solve this, because i cant keep my eyes open much longer

Answer 88

ok i think solving for n gives 96, obviously i'm wrong there or missing a step

Answer 89

because its 0.01 not 0.1........