1. Find limn→∞
n2 −6
.
2n2 +1
Finding the answer:
We look at the leading terms, in this case both numerator and denominator have a n2 so the
limit is given by its coefficients, 21 .
Now we have to use the definition to prove that rigorously. As you shall see, finding the
appropriate N for each is an art in itself and requires ability with inequalities (the triangle
inequality will be your dearest friend!).
In this case we need to work out the following expression
n2 − 6
1 2n2 − 12 − 2n2 − 1 13
− =
2
=
2
2n + 1 2
2(2n + 1)
2(2n2 + 1)
(1)
and we
qwant it to be less than , so that gives us a candidate for N , namely, any integer greater
− 12 .
than 13
4
Writing the answer:
(I am being extremely formal with the details here, but that won’t hurt you.)
We claim limn→∞
n2 −6
2n2 +1
= 21 .
lq
m
1
13
or
N
=
−
Indeed, let > 0 be fixed and define N = 1 if ≥ 13
otherwise 1 . Then
2
4
2
for n > N we have
n2 − 6
1 2n2 − 12 − 2n2 − 1 13
13
− =
=
<
2
2n + 1 2
2(2n2 + 1)
2(2n2 + 1)
2(2N 2 + 1)
q
13
,
the
choice
of
N
implies
N
>
− 12 , rearranging:
But, if < 13
2
4
13 1
−
< N2
4 2
1
2(2N 2 + 1)
<
13
13
< 2(2N 2 + 1)
Finally, note that if ≥
13
2
and
N = 1 the
last inequality also holds.
n2 −6
Therefore, if n > N then 2n2 +1 − 21 < as required. That is, limn→∞
1
dxe = the smallest integer greater than x.
1
n2 −6
2n2 +1
= 12 .