anonymous
  • anonymous
Show that (n-7)/(n+7) converges with the limit 1. When n goes from 1 to infinity.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Does it make sense to split it up into two: \[\frac{ n }{ n+7 }\] and \[\frac{ -7 }{ n+7 }\]
ganeshie8
  • ganeshie8
good idea, just try splitting it in another way
ganeshie8
  • ganeshie8
How about : \(\dfrac{n-7}{n+7} = \dfrac{(n+7)-14}{n+7} = 1 -\dfrac{14}{n+7}\)

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anonymous
  • anonymous
Yes, that is a great idea. Now i should be able to insert it in the definition for convergence.
ganeshie8
  • ganeshie8
Yes, remember \(\lim\limits_{n\to\infty} \dfrac{1}{n} = 0\)
ganeshie8
  • ganeshie8
wait a second, are you supposed to prove this using epsilon delta definition of limit ?
anonymous
  • anonymous
No, we have only learned the definition for convergence, and cauchy.
ganeshie8
  • ganeshie8
then we're good i hope :)
anonymous
  • anonymous
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: \[\frac{ 14 }{ n+7 }\le \frac{ 1 }{ n }\le \epsilon\]
anonymous
  • anonymous
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 \[n \ge N\]
ganeshie8
  • ganeshie8
so you do want to prove it using epsilong delta definition !
ganeshie8
  • ganeshie8
\[\frac{ 14 }{ n+7 }\le \frac{ 1 }{ n }\le \epsilon\] this is wrong, try again.
anonymous
  • anonymous
Ohh, that is epsilon delta definition. I didnt know that, all my courses and stuff is on another language.
anonymous
  • anonymous
Yea, i realize that, is there anyway that I can get to \[\frac{ 1 }{ n }\]
ganeshie8
  • ganeshie8
dont get attached to \(\dfrac{1}{n}\) so much
ganeshie8
  • ganeshie8
you cuold try somehting like this : \[\frac{ 14 }{ n+7 }\lt \frac{ 14 }{ n }\lt \epsilon\] so we want \(n\gt \dfrac{14}{\epsilon}\)
ganeshie8
  • ganeshie8
Now you're ready to start the proof
ganeshie8
  • ganeshie8
what we did so far is just the scratch work, its not proof ok
ganeshie8
  • ganeshie8
your proof must start with the value of \(N\) that works
anonymous
  • anonymous
Ohh, i thought that was the proof.
anonymous
  • anonymous
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 \[n >= N\]
anonymous
  • anonymous
I'm pretty sure, thats how they do in my books.
ganeshie8
  • ganeshie8
you're right, but you need to put that in proper order in your proof
ganeshie8
  • ganeshie8
if psble, may i see your complete proof ?
anonymous
  • anonymous
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
ganeshie8
  • ganeshie8
thats okay :) please take a look at 2nd example here : http://www.millersville.edu/~bikenaga/math-proof/limits-at-infinity/limits-at-infinity.html
ganeshie8
  • ganeshie8
your final proof must have that structure
anonymous
  • anonymous
I will look at it, thanks
ganeshie8
  • ganeshie8
np :)
anonymous
  • anonymous
At the start my book lets epsilon > 0, and then they do the "scratch work", and then they kinda concludes that it is convergent.
anonymous
  • anonymous
The only thing I havent seen / I dont understand is where they set M=max()
anonymous
  • anonymous
Hi
anonymous
  • anonymous
Hello
anonymous
  • anonymous
GM
anonymous
  • anonymous
@ganeshie8 When they use M, is that the N\[N \le n\] that I have been using?
ganeshie8
  • ganeshie8
Yes M is same as N
anonymous
  • anonymous
How old are yall
ganeshie8
  • ganeshie8
you may simply let \(M = \dfrac{14}{\epsilon}\) in our case forget about \(M = \max(0, \dfrac{14}{\epsilon})\) if it confuses you now..
anonymous
  • anonymous
How old are yall
ganeshie8
  • ganeshie8
for our probloem, it shouldn't matter either ways
anonymous
  • anonymous
Okay, thank you soo much! You are great :)
anonymous
  • anonymous
Okay I let y'all go
anonymous
  • anonymous
Bye

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