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the top and the bottom by n
2
:
x
n
=
1+1/n
2
1 − 1/n
2
=
a
n
b
n
,
where a
n
=1+
1
n
2
, b
n
=1−
1
n
2
.
Now, the limit law is applicable. We conclude that lim
n→∞
x
n
=1.
Later, when discussing the limit of a function, we will state a
theorem which deals with a limit of a rational function (that is, the
quotient of two polynomials).
Consider one other example. Let x
n
=
√
n(
√
n +1−
√
n). Again,
the limit law is inapplicable, we are in ∞·0 situation. A key word
here is “conjugate” (to try to get rid of one radical, at least). We
multiply x
n
by
1=
√
n +1+
√
n
√
n +1+
√
n
to get a new expression for the nth term:
x
n
=
√
n
√
n +1+
√
n
,
still indeterminate expression.
We now do something similar to the previous example: we divide
both the numerator and the denominator by
√
n to get an expression
where we conclude easily about the existence of the limit:
x
n
=
1
1+1/n +1
n→∞
−−−→
1
√
1+0+1
=
1
2
.
Definition 9
A sequence x
n
is bounded above if there is a number M such
that x
n
<M for all n 1.
It is bounded below if there is a number M such that x
n
>M for
all n 1.
6
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