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
