Indeterminate Forms - Florida State University
Indeterminate Forms
0 ,
,
0 ? ,
00,
0,
1,
-
0
These are the so called indeterminate forms.
One can apply L'Hopital's rule directly to the forms
0 0
and
.
It is simple to translate 0 ? into
0 1/
or into
1/0
,
for
example
one
can
write
limx xe-x
as
limx x/ex or as limx e-x/(1/x). To see that the exponent forms are indeterminate note that
ln 00 = 0 ln 0 = 0(-) = 0 ? , ln 0 = 0 ln = 0 ? , ln 1 = ln 1 = ? 0 = 0 ?
These formula's also suggest ways to compute these limits using L'Hopital's rule. Basically we use two
things, that ex and ln x are inverse functions of each other, and that they are continuous functions. If g(x)
is a continuous function then g(limxa f (x)) = limxa g(f (x)).
For
example
let's
figure
out
limx
(1
+
1 x
)x
=
e.
This
is
of
the
indeterminate
form
1.
We
write
exp(x)
for ex so to reduce the amount exponents.
lim (1 + 1 )x = exp(ln( lim (1 + 1 )x)) = exp( lim ln((1 + 1 )x))
x
x
x
x
x
x
= exp( lim
1 x ln(1 + )) = exp( lim
ln(1
+
1 x
)
)
x
x
x 1/x
We
can
now
apply
L'Hopital's
since
the
limit
is
of
the
form
0 0
.
=
exp( lim
x
(1/(1
+
1 x
))(-1/x2
-1/x2
)
)
=
exp( lim 1/(1
x
+
1 ))
x
=
exp(1)
=
e.
Exercises I. Find the limits.
A. lim (1 + 1 )3x B. lim (1 + k )x C. lim (1 + x)1/x D. lim xx E. lim x(x2) F. lim x1/ ln x.
x
x
x
x
x0
x0+
x0+
x0+
One might be tempted to handle - in a similar manner since
e-
=
e e
=
.
But L'Hopital's rule doesn't help here as the derivatives don't simplify. Instead, let f (x) and g(x) be
functions so that limxa f (x) = limxa g(x) = so that limxa(f (x) - g(x)) is - . One can rewrite
f (x)-g(x)
as
f (x)(1-g(x)/f (x)).
The
limit
limxa g(x)/f (x)
is
of
the
form
and
so
we
can
use
L'Hopital.
If limxa(1 - g(x)/f (x)) = c = 0 then limxa f (x)(1 - g(x)/f (x)) = c limxa f (x) =sign of c (sign of c is
? depending on the sign of c. On the otherhand, if c = 0, then f(x)(1-g(x)/f(x)) is of the form 0 ? which
we already know how to reexpress so that Hopital's rule can be applied.
For example let's show that limx( x + 1 - x) = 0. This is of the indeterminate form - .
x+1
limx( x + 1 -
x) = limx x(
x
- 1) = limx x(
1
(
1 + x - 1) = limx
1
+
1 x
-
1)
x-1/2
Now we can use L'Hopital's rule.
-1/x-2
2 1+1/x
2x3/2/x-2
=
limx (-1/2)x-3/2
=
limx 2
= limx
1 + 1/x
x
1 = 0.
1 + 1/x
Exercises II. Find the limits.
W. lim ((x + 1)3 - x3) X. lim (ln(x + 2) - ln(x)) Y. lim (3x - 2x) Z. lim (x-2 - x-1)
x
x
x
x0
................
................
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