3. What are Indeterminate Forms?
0
0 ¥
0 / 0 ¥ / ¥ 0 ×¥ ¥-¥
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
lim f (x) = 0
x®a
and
lim g(x) = 0
x®a
é f '(x) ù
if lim ê
ú exists, or¥ if this limit is + ¥ or - ¥, then
x®a ë g'(x) û
lim
x®a
f (x)
f '(x)
= lim
g(x) x®a g'(x)
5. How about some examples?
1.
2.
3.
4.
Find the following limits:
x2 - 4
2x
lim
= lim
= 2×2
x®2 x - 2
x®2 1
1- sin x
-cos x 0
lim
= lim
= =0
p
p
cos x
-1
x®
x® -sin x
2
2
e x -1
ex 1
lim 3 = lim 2 = = +¥
x®0
x®0 3x
x
0
7
1
4
4 -3
4 -3
- x
x
x 3
4
3
lim
= lim
= lim 3
= lim 1
=
x®+¥
x®+¥
x®+¥
x®+¥
æ1ö
æ1ö
æ1ö
1
æ1ö
3
sin ç ÷
- 2 cos ç ÷
cos ç ÷
3x cos ç ÷
èxø
èxø
èxø
x
èxø
4
æ 1 ö
3 (+¥) cos ç ÷
è +¥ ø
1
3
=
4
4
=
=0
(+¥) cos ( 0) +¥×1
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
lim f (x) = ¥
x®a
and
lim g(x) = ¥
x®a
é f '(x) ù
if lim ê
ú exists, or¥ if this limit is + ¥ or - ¥, then
x®a ë g'(x) û
lim
x®a
f (x)
f '(x)
= lim
g(x) x®a g'(x)
7. How about some examples?
Find the following limits:
x
1
1
1. lim x = lim x =
=0
x®+¥ e
x®+¥ e
+¥
1
ln x
sin x
sin x
x
lim
= lim
= lim
tan x = lim × limtan x =
2. x®0+
x®0+ -csc x cot x
x®0+ -x
x®0+
csc x
x x®0+
cos x
lim limtan x = (-1) × ( 0) = 0
+
x®0
1 x®0+
1
× cos x
ln (sin x )
cot x
1
sin x
lim
= lim
= lim
= lim
=
3. x®0+
x®0+
x®0+ cot x × sec 2 x
x®0+ sec 2 x
1
ln ( tan x )
× sec 2 x
tan x
lim cos2 x = cos2 0 = 1
+
x®0
8. Indeterminate Form of Type 0 ×¥
Can sometimes be evaluated by rewriting the product as
a ratio:
1
ln x
-x 2
= lim x = lim
= lim ( -x ) = 0
1. lim x ln x = lim
+
+
+
+
x®0
x®0
x®0
x®0+
1
1 x®0 x
- 2
x
x
2.
lim (1- tan x )
p
x®
(1- tan x ) = lim
sec 2x = lim
4
æp ö
sec ç ÷
2
è4ø
= =1
æ pö 2
2sin ç 2 × ÷
è 4ø
2
x®
p
4
cos2x
-sec 2 x
sec 2 x
= lim
=
p
p
x® -2sin 2x
x® -2sin 2x
4
4
9. Indeterminate Form of Type ¥-¥
Can sometimes be evaluated by combining the terms and
manipulating the result to produce quotient
1.
æ1
æ sin x - x ö
æ cos x -1 ö
1 ö
lim ç ÷ = lim ç
÷ = lim ç
÷=
+
+
+
x®0 è x
sin x ø x®0 è x sin x ø x®0 è sin x + x cos x ø
æ
ö
-sin x
0
lim ç
=
=0
÷
+
x®0 è cos x + cos x - x sin x ø
1+1- 0
æ ( e x -1) - x ö
æ x
ö
æ1
ö
1
e - x -1 ÷
2. lim
÷ = lim ç
=
ç - x ÷ = lim ç
x
x®0 x
x®0 ç x e -1 ÷
x®0 ç x e x -1 ÷
e -1 ø
è
) ø è ( )ø
è (
(e
x
-1)
ex
ex
1
1
lim x
= lim x x
= lim x
= lim
=
x®0 e -1 + xe x
x®0 e + e + xe x
x®0 e
( 2 + x ) x®0 x + 2 2
( )
10. ¥
Indeterminate Form of Type 0 ,¥ ,1
0
Can sometimes be evaluated by first introducing a
dependent variable
y = f (x)
g(x)
And then computing the limit of lny. Since
é f (x)g( x) ù = g(x)× ln [ f (x)]
ln y = ln ë
û
The limit of lny will be an indeterminate form of type 0 ×¥
0
11. How about an example?
Show that
y = (1+ x )
1
x
lim (1+ x ) = e
x®0
1
x
ln y = ln (1+ x )
1
x
ln (1+ x )
1
= ln (1+ x ) =
x
x
1
ln (1+ x )
lim ( ln y) = lim
= lim 1+ x = 1
x®0
x®0
x®0
x
1
lim ( ln y) =1
x®0
limy = e
1
x®0
1
x
lim (1+ x ) = e
x®0
