3. What are Indeterminate Forms?
¥/¥
0 / 0 0×¥ ¥-¥ 00
,¥0
,1¥
Applying L’Hopital’s Rule:
• Check that the limit of f(x) / g(x) is an indeterminate form 0 / 0 .
• Differentiate f and g separately.
• Find the limit of f’(x) / g’(x).
4. Indeterminate Form of Type 0 / 0
Suppose that f and g are differentiable functions on an open interval containing
x = a, except possibly at x = a, and that
if lim
x®a
f '(x)
g'(x)
é
ë
ê
ù
û
úexists, or¥ if this limit is +¥ or -¥, then
lim
x®a
f (x)= 0 and lim
x®a
g(x) = 0
lim
x®a
f (x)
g(x)
= lim
x®a
f '(x)
g'(x)
5. How about some examples?
Find the following limits:
lim
x®2
x2
- 4
x -2
= lim
x®2
2x
1
= 2×2
lim
x®
p
2
1-sin x
cosx
= lim
x®
p
2
-cosx
-sin x
=
0
-1
= 0
lim
x®0
ex
-1
x3
= lim
x®0
ex
3x2
=
1
0
= +¥
lim
x®+¥
x
-
4
3
sin
1
x
æ
è
ç
ö
ø
÷
= lim
x®+¥
-
4
3
x
-
7
3
-
1
x2
cos
1
x
æ
è
ç
ö
ø
÷
= lim
x®+¥
4
3
x
-
1
3
cos
1
x
æ
è
ç
ö
ø
÷
= lim
x®+¥
4
3x
1
3
cos
1
x
æ
è
ç
ö
ø
÷
=
4
3 +¥
( )
1
3 cos
1
+¥
æ
è
ç
ö
ø
÷
=
4
+¥
( )cos 0
( )
=
4
+¥×1
= 0
1.
2.
3.
4.
6. Indeterminate Form of Type
Suppose that f and g are differentiable functions on an open interval containing
x = a, except possibly at x = a, and that
if lim
x®a
f '(x)
g'(x)
é
ë
ê
ù
û
úexists, or¥ if this limit is +¥ or -¥, then
lim
x®a
f (x) =¥ and lim
x®a
g(x) =¥
lim
x®a
f (x)
g(x)
= lim
x®a
f '(x)
g'(x)
¥/¥
7. How about some examples?
Find the following limits:
lim
x®+¥
x
ex
= lim
x®+¥
1
ex
=
1
+¥
= 0
lim
x®0+
ln x
csc x
= lim
x®0+
1
x
-csc xcot x
= lim
x®0+
sin x
-x
tan x = lim
x®0+
-
sin x
x
× lim
x®0+
tan x =
lim
x®0+
-
cosx
1
lim
x®0+
tan x = -1
( )× 0
( ) = 0
lim
x®0+
ln sin x
( )
ln tan x
( )
= lim
x®0+
1
sin x
×cosx
1
tan x
×sec2
x
= lim
x®0+
cot x
cot x×sec2
x
= lim
x®0+
1
sec2
x
=
lim
x®0+
cos2
x = cos2
0 =1
1.
2.
3.
8. Indeterminate Form of Type 0×¥
Can sometimes be evaluated by rewriting the product as
a ratio:
lim
x®0+
xln x = lim
x®0+
ln x
1
x
= lim
x®0+
1
x
-
1
x2
= lim
x®0+
-x2
x
= lim
x®0+
-x
( )= 0
lim
x®
p
4
1- tan x
( )sec2x = lim
x®
p
4
1- tan x
( )
cos2x
= lim
x®
p
4
-sec2
x
-2sin2x
= lim
x®
p
4
sec2
x
-2sin2x
=
sec2 p
4
æ
è
ç
ö
ø
÷
2sin 2×
p
4
æ
è
ç
ö
ø
÷
=
2
2
=1
1.
2.
9. Indeterminate Form of Type ¥-¥
Can sometimes be evaluated by combining the terms and
manipulating the result to produce quotient
lim
x®0+
1
x
-
1
sin x
æ
è
ç
ö
ø
÷ = lim
x®0+
sin x - x
xsin x
æ
è
ç
ö
ø
÷ = lim
x®0+
cosx -1
sin x + xcosx
æ
è
ç
ö
ø
÷ =
lim
x®0+
-sin x
cosx +cosx - xsin x
æ
è
ç
ö
ø
÷ =
0
1+1- 0
= 0
lim
x®0
1
x
-
1
ex
-1
æ
è
ç
ö
ø
÷ = lim
x®0
ex
-1
( )- x
x ex
-1
( )
æ
è
ç
ç
ö
ø
÷
÷
= lim
x®0
ex
- x -1
x ex
-1
( )
æ
è
ç
ç
ö
ø
÷
÷
=
lim
x®0
ex
-1
( )
ex
-1
( )+ xex
= lim
x®0
ex
ex
+ex
+ xex
= lim
x®0
ex
ex
2+ x
( )
= lim
x®0
1
x +2
=
1
2
1.
2.
10. Indeterminate Form of Type
Can sometimes be evaluated by first introducing a
dependent variable
00
,¥0
,1¥
y = f (x)g(x)
And then computing the limit of lny. Since
lny = ln f (x)g(x)
é
ë ù
û= g(x)×ln f (x)
[ ]
The limit of lny will be an indeterminate form of type 0×¥
11. How about an example?
y = 1+ x
( )
1
x
ln y = ln 1+ x
( )
1
x =
1
x
ln 1+ x
( )=
ln 1+ x
( )
x
Show that lim
x®0
1+ x
( )
1
x = e
lim
x®0
ln y
( ) = lim
x®0
ln 1+ x
( )
x
= lim
x®0
1
1+ x
1
=1
lim
x®0
ln y
( )=1
x®0
limy = e1
lim
x®0
1+ x
( )
1
x = e
