1. prove that:
n²-1000n-> infinity as n-> infinity

We must show that given any M > 0,
there exists N > 0,
in terms of M, such that
when n > N, n²-1000n > M

To do this we work backwards first.  

We must find N in terms of M,
such that whenever n > N

n² - 1000n > M  

will always be true.

which means 

we must find a value of N, in terms of M,
such that whenever n > N

    n(n - 1000) > M

is always true whenever n > N

Now we can see that choosing
n > M+1000 will make this 
always true, since if

n > M+1000  

n(n - 1000) > (M+1000)(M+1000-1000) = M(M+1000) > M

So we take N = M+1000 and we have the preceding inequality. 

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2. use standard results to show that:
(n²-3n+1)/(n+5) ->infinity as n->infinity

We must show that given any M > 0, there exists 
N > 0, in terms of M, such that
when n > N, (n²-3n+1)/(n+5) > M

To do this we work backwards.  

We must find N in terms of M,
such that whenever n > N

(n²-3n+1)/(n+5) > M

This will be true whenever:

(n²-3n+1)/(n+5) - M > 0

and since n>0, this will be true
whenever

(n²-3n+1) - M(n+5) > 0
n² - 3n + 1 - Mn - 5M > 0

which will be true whenever

n² - (3+M)n + (1-5M) > 0

is true. 

Using the quadratic formula,
The left side has two zeros
        ________ 
(3+M ± Ö1+26M+M²)/2

and the inequality will
always be true when n is
greater than the larger
zero, which is
        ________ 
(3+M + Ö1+26M+M²)/2
  
So we take N as the larger
zero
             ________
 N = (3+M + Ö1+26M+M²)/2

then we can be sure that
when n > N, 
               
n > N, (n²-3n+1)/(n+5) > M

Edwin