Box "1/0"
More generally, we have the type L/0 here, where L is not zero
(zero over zero is a special type), but L can be infinite. Since we
can always factor out sign, we can assume that
L > 0.
Since we can also always do
L/0 = L⋅(1/0),
then the value of L will not influence the outcome, so it is enough to
know what to do with 1/0.
Assume that we want to find the limit of an expression
an/bn and we know that
an goes to 1 while bn
goes to 0.
Standard procedure: We try to find some N so that either
bn > 0 for
n > N (then bn
is of the type 0+ and we remember that
1/0+ = ∞);
or so that bn < 0 for
n > N (then
bn is of the type 0- and we remember that
1/0- = −∞).
If we are not able to find such N and bn
attains both signs for arbitrarily large N, then this particular
1/0 leads to a limit that does not exist.
Example:
Find the limit of
cn = 1/(2arctan(n) − π).
Solution: When we substitute infinity into cn,
we see that we face the type 1/0. What is the sign of the denominator for
n large? We always have
arctan(n) < π/2,
thus for every n
the denominator is negative. Therefore in our case we have
1/0-
and the answer is −∞.
Next box: indeterminate ratio
Back to Methods Survey
- Limits