Question

Show that (n-7)/(n+7) converges with the limit 1. When n goes from 1 to infinity.

Answer 1

Does it make sense to split it up into two:
\[\frac{ n }{ n+7 }\]
and
\[\frac{ -7 }{ n+7 }\]

Answer 2

good idea, just try splitting it in another way

Answer 3

How about :

\(\dfrac{n-7}{n+7} = \dfrac{(n+7)-14}{n+7} = 1 -\dfrac{14}{n+7}\)

Answer 4

Yes, that is a great idea. Now i should be able to insert it in the definition for convergence.

Answer 5

Yes, remember
\(\lim\limits_{n\to\infty} \dfrac{1}{n} = 0\)

Answer 6

wait a second, are you supposed to prove this using epsilon delta definition of limit ?

Answer 7

No, we have only learned the definition for convergence, and cauchy.

Answer 8

then we're good i hope :)

Answer 9

I insert it with the limit:\[|1-\frac{ 14 }{ n+7 }-1|=|-\frac{ 14 }{ n+7 }|=\frac{ 14 }{ n+7 }\]
Now this needs to be less or equal to epislon. Can I do the following:
\[\frac{ 14 }{ n+7 }\le \frac{ 1 }{ n }\le \epsilon\]

Answer 10

Because if i get it to \[\frac{ 1 }{ n }\], then i can isolate n:
\[n \ge \frac{ 1 }{ \epsilon }\]
And get that if a natural number N is bigger than \[\frac{ 1 }{ \epsilon }\] then the function is convergent for all \[n \ge N\]

Answer 11

so you do want to prove it using epsilong delta definition !

Answer 12

\[\frac{ 14 }{ n+7 }\le \frac{ 1 }{ n }\le \epsilon\]

this is wrong, try again.

Answer 13

Ohh, that is epsilon delta definition. I didnt know that, all my courses and stuff is on another language.

Answer 14

Yea, i realize that, is there anyway that I can get to \[\frac{ 1 }{ n }\]

Answer 15

dont get attached to \(\dfrac{1}{n}\) so much

Answer 16

you cuold try somehting like this :

\[\frac{ 14 }{ n+7 }\lt \frac{ 14 }{ n }\lt \epsilon\]

so we want \(n\gt \dfrac{14}{\epsilon}\)

Answer 17

Now you're ready to start the proof

Answer 18

what we did so far is just the scratch work, its not proof ok

Answer 19

your proof must start with the value of \(N\) that works

Answer 20

Ohh, i thought that was the proof.

Answer 21

Yea, but \[n>14/epsilon\] shows that for any number \[N>14/epsilon\], then \[|1-\frac{ 14 }{ n+7 }-1|\le \epsilon\] be true for all \[n >= N\]

Answer 22

I'm pretty sure, thats how they do in my books.

Answer 23

you're right, but you need to put that in proper order in your proof

Answer 24

if psble, may i see your complete proof ?

Answer 25

Well, the problem is that it is in Danish, and I am not great at english and dont know how to make a proof in english

Answer 26

thats okay :)
please take a look at 2nd example here :
http://www.millersville.edu/~bikenaga/math-proof/limits-at-infinity/limits-at-infinity.html

Answer 27

your final proof must have that structure

Answer 28

I will look at it, thanks

Answer 29

np :)

Answer 30

At the start my book lets epsilon > 0, and then they do the "scratch work", and then they kinda concludes that it is convergent.

Answer 31

The only thing I havent seen / I dont understand is where they set M=max()

Answer 32

Hi

Answer 33

Hello

Answer 34

GM

Answer 35

@ganeshie8 When they use M, is that the N\[N \le n\] that I have been using?

Answer 36

Yes M is same as N

Answer 37

How old are yall

Answer 38

you may simply let \(M = \dfrac{14}{\epsilon}\) in our case

forget about \(M = \max(0, \dfrac{14}{\epsilon})\) if it confuses you now..

Answer 39

How old are yall

Answer 40

for our probloem, it shouldn't matter either ways

Answer 41

Okay, thank you soo much! You are great :)

Answer 42

Okay I let y'all go

Answer 43

Bye