Proving Limit lim((n+2)/(n^2-3))=0 with Definition of a Limit

[mjjoga]

lim((n+2)/(n^2-3))=0 as n goes to infinity. I can only use the definition of a limit.

My work so far,

I'm trying to work out what n will be greater than. I have:
|(n+2)/(n^2-3)|=(n+2)/(n^2-3) if n is greater than one.
From here, I have been trying anything to get rid of the addition and subtraction.
I got 1/(n-2) but that can't work. I'm not sure how to manipulate it. I tried multiplying by n, making perfect squares. I'm stuck. If I could get a hint, that would be great.
mjjoga

 

[Char. Limit]

|(n+2)/(n^2-3)|=(n+2)/(n^2-3) if n is greater than one.

I tried plugging in n=1.1 and got a positive number on the right and a negative number on the left. I think you mean if n>sqrt(3), unless you're only referring to integers n.

 

[tiny-tim]

hi mjjoga!

(try using the X² icon just above the Reply box )

can you do lim (n + 2)/(n² - 4) ?

 

[mjjoga]

you know what, when I was canceling out the n+2 on the top, I had put down ^{(n+2)2 in the denominator (I tried to use the x^2 button) so I left x+2 on the bottom, but it's the sum and difference so I need n-2 on the bottom. then it's fine. Sometimes my algebra just flies out the window...
Thank you bunches,
mjjoga
ps-I might be back, I've got 5 problems left...}

 

[mjjoga]

n is in natural numbers

 

[mjjoga]

Define N>ε+2. Then whenever n>max(N,2), we have |(n+2)/(n^2-3)|<(n+2)/(n^2-3)<(n+2)/(n^2-4)=1/(n-2)<1/(ε+2-2)=1/ε
1/epsilon isn't less than epsilon. I don't know what I'm doing wrong. i added that n>2 so that the denumerator does not equal 0. I know that 1/(n-2) is right, but what else can i do to it

 

[tiny-tim]

Define N>ε+2.

erm …

ε is very small, but N is very large …

wouldn't you be better off using 1/ε ?

 

[mjjoga]

hm, but epsilon could be a tiny fraction and make 1/epsilon really big. There must be some sort of trick we can use to get it less than epsilon.

 

[tiny-tim]

… to get it less than epsilon.

you don't need N < ε, you need an M such that if N > M, then the whole thing is < ε …

your ε must be very small, and your N (or M) must be very large

 

[mjjoga]

I figured it out. If I multiply by -1/-1 then with the different signs, I can get rid of the addition and subtraction. I did this before I even factored. then it simplifies to 1/n, so I let N>1/epsilon and it works.

 

[tiny-tim]

thanks for the help! by the way, if I know that n is greater than 1 and is a natural number, can i make a fraction bigger by multiplying the denominator by sqrt(n)?

hi myriam!

yes …

the simple rule is that if you make the _denominator bigger, you make the whole thing smaller (and vice versa)