thirteen

Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
Indeterminate Forms and L’Hospital’s Rule
THEOREM (L’Hospital’s Rule): Suppose f and g are differentiable and g?(x) 6= 0 near a
(except possibly at a). Suppose that
lim
x?a
f(x) = 0 and lim
x?a
g(x) = 0
or that
lim
x?a
f(x) = ±? and lim
x?a
g(x) = ±?
(In other words, we have an indeterminate form of type
0
0
or ?
?
.) Then
lim
x?a
f(x)
g(x)
= lim
x?a
f?(x)
g?(x)
if the limit on the right side exists (or is ? or ??).
Indeterminate Forms of Type
0
0
and ?
?
EXAMPLES:
1. Find lim
x??
5x ? 2
7x + 3
.
Solution 1: We have
lim
x??
5x ? 2
7x + 3
=
h
?
?
i
= lim
x??
5x?2
x
7x+3
x
= lim
x??
5x
x ? 2
x
7x
x + 3
x
= lim
x??
5 ? 2
x
7 + 3
x
=
5 ? 0
7 + 0
=
5
7
Solution 2: We have
lim
x??
5x ? 2
7x + 3
=
h
?
?
i
= lim
x??
(5x ? 2)?
(7x + 3)? = lim
x??
(5x)? ? 2?
(7x)? + 3? = lim
x??
5x? ? 2?
7x? + 3?
= lim
x??
5 · 1 ? 0
7 · 1 + 0
= lim
x??
5
7
=
5
7
In short,
lim
x??
5x ? 2
7x + 3
= lim
x??
(5x ? 2)?
(7x + 3)? = lim
x??
5
7
=
5
7
2. Find lim
x??
x5 + x4 + x3 + x2 + x + 1
2x5 + x4 + x3 + x2 + x + 1
.
1
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
2. Find lim
x??
x5 + x4 + x3 + x2 + x + 1
2x5 + x4 + x3 + x2 + x + 1
.
Solution 1: We have
lim
x??
x5 + x4 + x3 + x2 + x + 1
2x5 + x4 + x3 + x2 + x + 1
=
h
?
?
i
= lim
x??
x5+x4+x3+x2+x+1
x5
2x5+x4+x3+x2+x+1
x5
= lim
x??
x5
x5 + x4
x5 + x3
x5 + x2
x5 + x
x5 + 1
x5
2x5
x5 + x4
x5 + x3
x5 + x2
x5 + x
x5 + 1
x5
= lim
x??
1 + 1
x + 1
x2 + 1
x3 + 1
x4 + 1
x5
2 + 1
x + 1
x2 + 1
x3 + 1
x4 + 1
x5
=
1 + 0 + 0 + 0 + 0 + 0
2 + 0 + 0 + 0 + 0 + 0
=
1
2
Solution 2: We have
lim
x??
x5 + x4 + x3 + x2 + x + 1
2x5 + x4 + x3 + x2 + x + 1
=
h
?
?
i
= lim
x??
(x5 + x4 + x3 + x2 + x + 1)?
(2x5 + x4 + x3 + x2 + x + 1)?
= lim
x??
5x4 + 4x3 + 3x2 + 2x + 1
10x4 + 4x3 + 3x2 + 2x + 1
=
h
?
?
i
= lim
x??
(5x4 + 4x3 + 3x2 + 2x + 1)?
(10x4 + 4x3 + 3x2 + 2x + 1)?
= lim
x??
20x3 + 12x2 + 6x + 2
40x3 + 12x2 + 6x + 2
=
h
?
?
i
= lim
x??
(20x3 + 12x2 + 6x + 2)?
(40x3 + 12x2 + 6x + 2)?
= lim
x??
60x2 + 24x + 6
120x2 + 24x + 6
=
h
?
?
i
= lim
x??
(60x2 + 24x + 6)?
(120x2 + 24x + 6)?
= lim
x??
120x + 24
240x + 24
=
h
?
?
i
= lim
x??
(120x + 24)?
(240x + 24)? = lim
x??
120
240
=
120
240
=
1
2
3. Find lim
x??2
x + 2
ln(x + 3)
.
Solution: We have
lim
x??2
x + 2
ln(x + 3)
=

0
0

= lim
x??2
(x + 2)?
(ln(x + 3))? =
???
??
lim
x??2
(x + 2)?
1
x + 3 · (x + 3)?
= lim
x??2
x? + 2?
1
x + 3 · (x? + 3?)
= lim
x??2
1 + 0
1
x + 3 · (1 + 0)
???
??
= lim
x??2
1
1
x + 3
= lim
x??2
1 · (x + 3)
1
x + 3 · (x + 3)
= lim
x??2
x + 3
1
= lim
x??2
(x + 3)
= ?2 + 3 = 1
In short,
lim
x??2
x + 2
ln(x + 3)
= lim
x??2
(x + 2)?
(ln(x + 3))? = lim
x??2
1
(x + 3)?1 = lim
x??2
(x + 3) = 1
4. Find lim
x??
3x
x2 + x ? 1
.
2
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
4. Find lim
x??
3x
x2 + x ? 1
.
Solution: We have
lim
x??
3x
x2 + x ? 1
=
h
?
?
i
= lim
x??
(3x)?
(x2 + x ? 1)? = lim
x??
3x ln 3
2x + 1
=
h
?
?
i
= lim
x??
(3x ln 3)?
(2x + 1)?
= lim
x??
ln 3 (3x)?
2
= lim
x??
ln 3 · 3x · ln 3
2
= ? (D.N.E.)
5. Find lim
x??
ln x
?x
.
Solution: We have
lim
x??
ln x
?x
=
h
?
?
i
= lim
x??
(ln x)?
(x1/2)? = lim
x??
x?1
1
2x?1/2
= lim
x??
x?1 · x
1
2x?1/2 · x
= lim
x??
1
1
2x1/2
= 0
6. Find lim
x?0
5x ? tan 5x
x3
.
Solution 1: We have
lim
x?0
5x ? tan 5x
x3 =

0
0

= lim
x?0
(5x ? tan 5x)?
(x3)? = lim
x?0
5 ? sec2 5x · (5x)?
3x2 = lim
x?0
5 ? sec2 5x · 5
3x2
= lim
x?0
5(1 ? sec2 5x)
3x2 =
5
3
lim
x?0
1 ? sec2 5x
x2
Since lim
x?0
1 ? sec2 5x
x2 is an indeterminate form of type
0
0
, we can use L’Hospital’s Rule again.
But it is easier to do trigonometry instead. Note that 1 ? sec2 5x = ?tan2 5x. Therefore
5
3
lim
x?0
1 ? sec2 5x
x2 =
5
3
lim
x?0
?tan2 5x
x2 = ?
5
3
lim
x?0
tan2 5x
x2 = ?
5
3
lim
x?0
sin2 5x
cos2 5x
x2 = ?
5
3
lim
x?0
sin2 5x
1
x2
= ?
5
3
lim
x?0
sin2 5x
x2 =

0
0

= ?
5
3
lim
x?0

sin 5x
x
2
= ?
5
3
lim
x?0

5 ·
sin 5x
5x
2
= ?
5
3

5 lim
x?0
sin 5x
5x
2
= ?
5
3

5 lim
5x?0
sin 5x
5x
2
=

lim
u?0
sin u
u
= 1

= ?
5
3
(5 · 1)2 = ?
5
3 · 25 = ?
125
3
In short,
lim
x?0
5x ? tan 5x
x3 = lim
x?0
(5x ? tan 5x)?
(x3)? =
5
3
lim
x?0
1 ? sec2 5x
x2 = ?
5
3
lim
x?0
tan2 5x
x2 = ?
5
3
lim
x?0
sin2 5x
x2
= ?
5
3
lim
x?0

5 ·
sin 5x
5x
2
= ?
5
3
(5 · 1)2 = ?
125
3
Solution 2??? (WRONG!): We have
lim
x?0
5x ? tan 5x
x3 =

0
0

= lim
x?0
5x ?
sin 5x
cos 5x
x3
??? = lim
x?0
5x ?
sin 5x
1
x3 = lim
x?0
5x ? sin 5x
x3 =

0
0

= lim
x?0
(5x ? sin 5x)?
(x3)? = lim
x?0
5 ? 5 cos 5x
3x2 =

0
0

= lim
x?0
(5 ? 5 cos 5x)?
(3x2)? = lim
x?0
25 sin 5x
6x
=
125
6
3
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
7. Find lim
x??
x + sin x
x + cos x
.
Solution 1: We have
lim
x??
x + sin x
x + cos x
=
h
?
?
i
= lim
x??
x+sin x
x
x+cos x
x
= lim
x??
x
x + sin x
x
x
x + cos x
x
= lim
x??
1 + sin x
x
1 + cos x
x
It is easy to show that lim
x??
sin x
x
= 0 and lim
x??
cos x
x
= 0 by the Squeeze Theorem. Therefore
lim
x??
x + sin x
x + cos x
= lim
x??
1 + sin x
x
1 + cos x
x
=
1 + 0
1 + 0
= 1
Solution 2(???): We have
lim
x??
x + sin x
x + cos x
=
h
?
?
i
= lim
x??
(x + sin x)?
(x + cos x)? = lim
x??
x? + (sin x)?
x? + (cos x)? = lim
x??
1 + cos x
1 ? sin x
One can show, however, that lim
x??
1 + cos x
1 ? sin x
does not exist. In fact, we first note that 1 + cos x
and 1 ? sin x may attain any value between 0 and 2. From this one can deduce that
1 + cos x
1 ? sin x
attains any nonnegative value infinitely often as x ? ?. This means that lim
x??
1 + cos x
1 ? sin x
does
not exist, so L’Hospital’s Rule can’t be applied here.
8. Find lim
x??
sin x
1 ? cos x
.
Solution(???): We have
lim
x??
sin x
1 ? cos x
= lim
x??
(sin x)?
(1 ? cos x)? = lim
x??
cos x
sin x
= ??
This is WRONG. In fact, although the numerator sin x ? 0 as x ? ?, notice that the
denominator (1 ? cos x) does not approach 0, so L’Hospital’s Rule can’t be applied here. The
required limit is easy to find, because the function is continuous at  and the denominator is
nonzero here:
lim
x??
sin x
1 ? cos x
=
sin 
1 ? cos 
=
0
1 ? (?1)
= 0
9. Find lim
x?0
sin x
x
.
Solution(???): We have
lim
x?0
sin x
x
=

0
0

= lim
x?0
(sin x)?
x? = lim
x?0
cos x
1
=
cos 0
1
=
1
1
= 1
The answer is correct, but the solution is WRONG. Indeed, the above proof is based on the
formula (sin x)? = cos x. But this result was deduced from the fact that lim
x?0
sin x
x
= 1 (see
Appendix A). So, the solution is wrong because it is based on Circular Reasoning which is a
logical fallacy. However, one can apply L’Hospital’s Rule to modifications of this limit (see
Appendix B).
4
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
Indeterminate Forms of Type ??? and 0 ·?
EXAMPLES:
10. Find lim
x??
(x ? ln x).
Solution 1: We have
lim
x??
(x ? ln x) = [???] = lim
x??

x · 1 ? x ·
ln x
x

= lim
x??
x

1 ?
ln x
x

Note that
lim
x??
ln x
x
=
h
?
?
i
= lim
x??
(ln x)?
x? = lim
x??
x?1
1
= lim
x??
1
x
= 0
therefore
lim
x??
(x ? ln x) = lim
x??
x

1 ?
ln x
x

= lim
x??
x (1 ? 0) = ? (D.N.E.)
Solution 2: We have
lim
x??
(x ? ln x) = [???] = lim
x??
(ln(ex) ? ln x) = lim
x??
ln

ex
x

= ln

lim
x??
ex
x

=
h
?
?
i
= ln

lim
x??
(ex)?
x?

= ln

lim
x??
ex
1

= ? (D.N.E.)
11. Find lim
x?1

1
ln x ?
1
x ? 1

.
5
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
11. Find lim
x?1

1
ln x ?
1
x ? 1

.
Solution: We have
lim
x?1

1
ln x ?
1
x ? 1

= [???] = lim
x?1

1 · (x ? 1)
ln x · (x ? 1) ?
ln x · 1
ln x · (x ? 1)

= lim
x?1
x ? 1 ? ln x
ln x(x ? 1)
=

0
0

= lim
x?1
(x ? 1 ? ln x)?
(ln x(x ? 1))?
= lim
x?1
x? ? 1? ? (ln x)?
(ln x)? · (x ? 1) + ln x · (x ? 1)? = lim
x?1
1 ?
1
x
x ? 1
x
+ ln x
= lim
x?1

1 ?
1
x

x

x ? 1
x
+ ln x

x
= lim
x?1
1 · x ?
1
x · x
x ? 1
x · x + ln x · x
= lim
x?1
x ? 1
x ? 1 + x ln x
=

0
0

= lim
x?1
(x ? 1)?
(x ? 1 + x ln x)? = lim
x?1
x? ? 1?
x? ? 1? + x? ln x + x(ln x)?
= lim
x?1
1 ? 0
1 ? 0 + 1 · ln x + x ·
1
x
= lim
x?1
1
2 + ln x
=
1
2 + 0
=
1
2
In short,
lim
x?1

1
ln x ?
1
x ? 1

= lim
x?1
x ? 1 ? ln x
ln x · (x ? 1)
= lim
x?1
(x ? 1 ? ln x)?
(ln x · (x ? 1))? = lim
x?1
1 ?
1
x
x ? 1
x
+ ln x
= lim
x?1
x ? 1
x ? 1 + x ln x
= lim
x?1
(x ? 1)?
(x ? 1 + x ln x)? = lim
x?1
1
2 + ln x
=
1
2
12. Find lim
x?0+
(sin x ln x).
6
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
12. Find lim
x?0+
(sin x ln x).
Solution 1: We have
lim
x?0+
(sin x ln x) = [0 ·?] =

lim
x?0+
sin x ln x
1
= lim
x?0+
(sin x)?1 · sin x ln x
(sin x)?1 · 1

= lim
x?0+
ln x
(sin x)?1 =
h
?
?
i
= lim
x?0+
(ln x)?
((sin x)?1)? = lim
x?0+
x?1
?(sin x)?2 · (sin x)? = lim
x?0+
x?1
?(sin x)?2 cos x
= lim
x?0+
x?1 · x sin2 x
?(sin x)?2 cos x · x sin2 x
= ? lim
x?0+
sin2 x
x cos x
We can now proceed in two different ways. Either
? lim
x?0+
sin2 x
x cos x
=

0
0

= ? lim
x?0+

sin x
x ·
sin x
cos x

= ? lim
x?0+
sin x
x · lim
x?0+
sin x
cos x
= ?1 ·
sin 0
cos 0
= ?1 ·
0
1
= 0
or
? lim
x?0+
sin2 x
x cos x
=

0
0

= ? lim
x?0+
(sin2 x)?
(x cos x)? = ? lim
x?0+
2 sin x cos x
x? cos x + x(cos x)?
= ? lim
x?0+
2 sin x cos x
cos x ? x sin x
= ?
2 sin 0 cos 0
cos 0 ? 0 · sin 0
= ?
2 · 0 · 1
1 ? 0 · 0
= 0
In short,
lim
x?0+
(sin x ln x) = lim
x?0+
ln x
(sin x)?1 = lim
x?0+
(ln x)?
((sin x)?1)? = lim
x?0+
x?1
?(sin x)?2 cos x
= ? lim
x?0+
sin2 x
x cos x
= ? lim
x?0+

sin x
x ·
sin x
cos x

= ?1 · 0 = 0
Solution 2: We have
lim
x?0+
(sin x ln x) = [0 ·?] =

lim
x?0+
sin x ln x
1
= lim
x?0+
sin x ln x · ln?1 x
1 · ln?1 x

= lim
x?0+
sin x
ln?1 x
=

0
0

= lim
x?0+
(sin x)?
??
ln?1 x

? = lim
x?0+
cos x
?ln?2 x · (ln x)? = lim
x?0+
cos x
?ln?2 x · x?1
= lim
x?0+
ln2 x · cos x
?ln2 x · ln?2 x · x?1
= lim
x?0+
ln2 x · cos x
?x?1 = lim
x?0+
ln2 x
?x?1 · lim
x?0+
cos x = lim
x?0+
ln2 x
?x?1 =
h
?
?
i
= lim
x?0+
??
ln2 x

?
(?x?1)?
= lim
x?0+
2 ln x · (ln x)?
x?2 = lim
x?0+
2 ln x · x?1
x?2 = lim
x?0+
2 ln x · x?1 · x
x?2 · x
= lim
x?0+
2 ln x
x?1 =
h
?
?
i
= lim
x?0+
(2 ln x)?
(x?1)? = lim
x?0+
2x?1
?x?2 = lim
x?0+
2x?1 · x2
?x?2 · x2 = lim
x?0+
2x
?1
=
2 · 0
?1
= 0
7
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
Indeterminate Forms of Type ?0, 00 and 1?
13. Find lim
x??
x1/x.
Solution (version 1): Note that lim
x??
x1/x is ?0 type of an indeterminate form. Put
y = x1/x
then
ln y = ln x1/x =
1
x
ln x =
ln x
x
We have
lim
x??
ln x
x
=
h
?
?
i
= lim
x??
(ln x)?
x? = lim
x??
x?1
1
= lim
x??
1
x
= 0
Therefore
lim
x??
x1/x = e0 = 1
Solution (version 2): We have
lim
x??
x1/x = lim
x??
eln x1/x
= lim
x??
e
ln x
x = e
lim
x??
ln x
x = e
lim
x??
(ln x)?
x? = e
lim
x??
x?1
1 = e
lim
x??
1
x = e0 = 1
14. Find lim
x?/2
(tan x)2x?.
8
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
14. Find lim
x?/2
(tan x)2x?.
Solution: Note that lim
x?/2
(tan x)2x? is ?0 type of an indeterminate form. Put
y = (tan x)2x?
then
ln y = ln
??
(tan x)2x?

= (2x ? ) ln(tan x)
=

(2x ? ) ln(tan x)
1
=
(2x ? )?1 · (2x ? ) ln(tan x)
(2x ? )?1 · 1

=
ln(tan x)
(2x ? )?1
We have
lim
x?/2
ln(tan x)
(2x ? )?1 =
h
?
?
i
= lim
x?/2
[ln(tan x)]?
[(2x ? )?1]? = lim
x?/2
1
tan x · (tan x)?
(?1)(2x ? )?2 · (2x ? )?
= lim
x?/2
1
tan x · sec2 x
(?1)(2x ? )?2 · 2
= ?
1
2
lim
x?/2
1
sin x
cos x · 1
cos2 x
(2x ? )?2 = ?
1
2
lim
x?/2
1
sin x
cos x
·cos2 x
(2x ? )?2
= ?
1
2
lim
x?/2
1
sin x cos x
(2x ? )?2 = ?
1
2
lim
x?/2
1
sin x cos x · sin x cos x(2x ? )2
(2x ? )?2 · sin x cos x(2x ? )2 = ?
1
2
lim
x?/2
(2x ? )2
sin x cos x
= ?
1
2
lim
x?/2
(2x ? )2
sin
??
2


cos x
= ?
1
2
lim
x?/2
(2x ? )2
cos x
=

0
0

= ?
1
2
lim
x?/2
[(2x ? )2]?
(cos x)?
= ?
1
2
lim
x?/2
2(2x ? ) · (2x ? )?
?sin x
= ?
1
2
lim
x?/2
2(2x ? ) · 2
?sin x
= ?
1
2 ·
2 ·

2 ·

2 ? 

· 2
?sin

2
= ?
1
2 ·
2 · 0 · 2
?1
= 0
Therefore
lim
x?/2
(tan x)2x? = e0 = 1
In short,
y = (tan x)2x? =? ln y = (2x ? ) ln(tan x) =
ln(tan x)
(2x ? )?1
We have
lim
x?/2
ln(tan x)
(2x ? )?1 = lim
x?/2
[ln(tan x)]?
[(2x ? )?1]? = lim
x?/2
1
tan x · sec2 x
(?1)(2x ? )?2 · 2
= ?
1
2
lim
x?/2
(2x ? )2
sin x cos x
= ?
1
2
lim
x?/2
(2x ? )2
cos x
=

0
0

= ?
1
2
lim
x?/2
[(2x ? )2]
?
(cos x)?
= ?
1
2
lim
x?/2
2(2x ? ) · 2
?sin x
= ?
1
2 ·
2 ·

2 ·

2 ? 

· 2
?sin

2
= 0
Therefore lim
x?/2
(tan x)2x? = e0 = 1.
9
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
15. Find lim
x?0+
xx.
Solution: Note that lim
x?0+
xx is 00 type of an indeterminate form. Put
y = xx
then
ln y = ln xx = x ln x =

x ln x
1
=
x?1 · x ln x
x?1 · 1

=
ln x
x?1
We have
lim
x?0+
ln x
x?1 =
h
?
?
i
= lim
x?0+
(ln x)?
(x?1)? = lim
x?0+
x?1
?x?2
= lim
x?0+
x?1 · x2
?x?2 · x2 = lim
x?0+
x
?1
=
0
?1
= 0
Therefore
lim
x?0+
xx = e0 = 1
In short,
y = xx =? ln y = x ln x =
ln x
x?1
We have
lim
x?0+
ln x
x?1 = lim
x?0+
(ln x)?
(x?1)? = lim
x?0+
x?1
?x?2 = ? lim
x?0+
x = 0
Therefore
lim
x?0+
xx = e0 = 1
16. Find lim
x?0+
(tan 5x)x.
10
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
16. Find lim
x?0+
(tan 5x)x.
Solution: Note that lim
x?0+
(tan 5x)x is 00 type of an indeterminate form. Put
y = (tan 5x)x
then
ln y = ln(tan 5x)x = x ln(tan 5x) =

x ln(tan 5x)
1
=
x?1 · x ln(tan 5x)
x?1 · 1

=
ln(tan 5x)
x?1
We have
lim
x?0+
ln(tan 5x)
x?1 =
h
?
?
i
= lim
x?0+
(ln(tan 5x))?
(x?1)? = lim
x?0+
1
tan 5x · (tan 5x)?
?x?2
= lim
x?0+
1
tan 5x · sec2 5x · (5x)?
?x?2 = lim
x?0+
1
tan 5x · sec2 5x · 5
?x?2
= lim
x?0+
1
sin 5x
cos 5x · sec2 5x · 5
?x?2 = lim
x?0+
1
sin 5x
cos 0 · sec2 0 · 5
?x?2
= lim
x?0+
1
sin 5x · 5
?x?2 = lim
x?0+
1
sin 5x · 5 · x2 sin 5x
?x?2 · x2 sin 5x
= lim
x?0+
5x2
?sin 5x
=

0
0

= lim
x?0+
(5x2)?
(?sin 5x)?
= lim
x?0+
10x
?cos 5x · (5x)? = lim
x?0+
10x
?cos 5x · 5
=
10 · 0
?cos 0 · 5
= 0
Therefore
lim
x?0+
(tan 5x)x = e0 = 1
In short,
y = (tan 5x)x =? ln y = x ln(tan 5x) =
ln(tan 5x)
x?1
We have
lim
x?0+
ln(tan 5x)
x?1 = lim
x?0+
(ln(tan 5x))?
(x?1)? = lim
x?0+
1
tan 5x · sec2 5x · 5
?x?2 = lim
x?0+
1
sin 5x
cos 5x · sec2 5x · 5
?x?2
= lim
x?0+
1
sin 5x · 5
?x?2 = lim
x?0+
5x2
?sin 5x
= lim
x?0+
(5x2)?
(?sin 5x)? = lim
x?0+
10x
?cos 5x · 5
= 0
Therefore
lim
x?0+
(tan 5x)x = e0 = 1
11
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
17. Find lim
x?0+
(sin 2x)tan 3x.
Solution: Note that lim
x?0+
(sin 2x)tan 3x is 00 type of an indeterminate form. Put
y = (sin 2x)tan 3x
then
ln y = ln
??
(sin 2x)tan 3x

= tan 3x ln(sin 2x) =
sin 3x ln(sin 2x)
cos 3x
We have
lim
x?0+
sin 3x ln(sin 2x)
cos 3x
=
lim
x?0+
(sin 3x ln(sin 2x))
lim
x?0+
cos 3x
= lim
x?0+
(sin 3x ln(sin 2x)) = [0 ·?]
=

lim
x?0+
sin 3x ln(sin 2x)
1
= lim
x?0+
(sin 3x)?1 · sin 3x ln(sin 2x)
(sin 3x)?1 · 1

= lim
x?0+
ln(sin 2x)
(sin 3x)?1 =
h
?
?
i
= lim
x?0+
[ln(sin 2x)]?
((sin 3x)?1)? = lim
x?0+
sin?1 2x · (sin 2x)?
?(sin 3x)?2 · (sin 3x)? = lim
x?0+
sin?1 2x · cos 2x · (2x)?
?(sin 3x)?2 · cos 3x · (3x)?
= lim
x?0+
sin?1 2x · cos 2x · 2
?(sin 3x)?2 · cos 3x · 3
= ?
2
3
lim
x?0+
sin?1 2x
(sin 3x)?2 · lim
x?0+
cos 2x
cos 3x
= ?
2
3
lim
x?0+
sin?1 2x
(sin 3x)?2
= ?
2
3
lim
x?0+
sin?1 2x · sin 2x · sin2 3x
(sin 3x)?2 · sin 2x · sin2 3x
= ?
2
3
lim
x?0+
sin2 3x
sin 2x
=

0
0

= ?
2
3
lim
x?0+
(sin2 3x)?
(sin 2x)?
= ?
2
3
lim
x?0+
2 sin 3x · (sin 3x)?
cos 2x · (2x)? = ?
2
3
lim
x?0+
2 sin 3x · cos 3x · (3x)?
cos 2x · 2
= ?
2
3
lim
x?0+
2 sin 3x · cos 3x · 3
cos 2x · 2
= ?
2
3 ·
2 sin 0 · cos 0 · 3
cos 0 · 2
= ?
2
3 ·
2 · 0 · 1 · 3
1 · 2
= 0
Therefore
lim
x?0+
(sin 2x)tan 3x = e0 = 1
In short,
y = (sin 2x)tan 3x =? ln y = tan 3x ln(sin 2x) =
sin 3x ln(sin 2x)
cos 3x
We have
lim
x?0+
sin 3x ln(sin 2x)
cos 3x
= lim
x?0+
(sin 3x ln(sin 2x)) = lim
x?0+
ln(sin 2x)
(sin 3x)?1 = lim
x?0+
[ln(sin 2x)]?
((sin 3x)?1)?
= lim
x?0+
sin?1 2x · cos 2x · 2
?(sin 3x)?2 · cos 3x · 3
= ?
2
3
lim
x?0+
sin?1 2x
(sin 3x)?2 = ?
2
3
lim
x?0+
sin2 3x
sin 2x
= ?
2
3
lim
x?0+
(sin2 3x)?
(sin 2x)?
= ?
2
3
lim
x?0+
2 sin 3x · cos 3x · 3
cos 2x · 2
= ?
2
3 ·
2 sin 0 · cos 0 · 3
cos 0 · 2
= 0
Therefore lim
x?0+
(sin 2x)tan 3x = e0 = 1.
12
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
18. Find lim
x??

x + 1
x + 2
x
.
Solution 1: Note that lim
x??

x + 1
x + 2
x
is 1? type of an indeterminate form. Put y =

x + 1
x + 2
x
,
then
ln y = ln

x + 1
x + 2
x
= x ln

x + 1
x + 2

=
????
???
x ln

x + 1
x + 2

1
=
x?1 · x ln

x + 1
x + 2

x?1 · 1
????
???
=
ln

x + 1
x + 2

x?1
We have
lim
x??
ln

x + 1
x + 2

x?1 =

0
0

= lim
x??

ln

x + 1
x + 2
?
(x?1)? = lim
x??
1
x+1
x+2 ·

x + 1
x + 2
?
?x?2
= lim
x??
x + 2
x + 1 ·
(x + 1)?(x + 2) ? (x + 1)(x + 2)?
(x + 2)2
?x?2 = lim
x??
x + 2
x + 1 ·
1 · (x + 2) ? 1 · (x + 1)
(x + 2)2
?x?2
= lim
x??
x + 2
x + 1 ·
x + 2 ? x ? 1
(x + 2)2
?x?2 = lim
x??
x + 2
x + 1 ·
1
(x + 2)2
?x?2 = lim
x??
1
(x + 1)(x + 2)
?x?2
= lim
x??
1
(x + 1)(x + 2) · x2(x + 1)(x + 2)
?x?2 · x2(x + 1)(x + 2)
= lim
x??
x2
?(x + 1)(x + 2)
=
h
?
?
i
= ? lim
x??
x2
x2 + 3x + 1
= ? lim
x??
x2
x2
x2 + 3x + 1
x2
= ? lim
x??
x2
x2
x2
x2 +
3x
x2 +
1
x2
= ? lim
x??
1
1 +
3
x
+
1
x2
= ?
1
1 + 0 + 0
= ?1
Therefore
lim
x??

x + 1
x + 2
x
= e?1
In short,
y =

x + 1
x + 2
x
=? ln y = x ln

x + 1
x + 2

=
ln

x + 1
x + 2

x?1
We have
lim
x??
ln

x + 1
x + 2

x?1 = lim
x??

ln

x + 1
x + 2
?
(x?1)? = lim
x??
x + 2
x + 1 ·
1
(x + 2)2
?x?2 = ? lim
x??
x2
(x + 1)(x + 2)
= ? lim
x??
x2
x2 + 3x + 1
= ?1
Therefore lim
x??

x + 1
x + 2
x
= e?1.
13
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
Solution 2: We have
lim
x??

x + 1
x + 2
x
= lim
x??

x + 2 ? 1
x + 2
x
= lim
x??

x + 2
x + 2 ?
1
x + 2
x
= lim
x??

1 ?
1
x + 2
x
= lim
x??

1 +
1
?x ? 2
(?x?2) x
?x?2
= lim
x??
"
1 +
1
?x ? 2
?x?2
# x
?x?2
Since lim
u?±?

1 +
1
u
u
= e, it follows that lim
x??

1 +
1
?x ? 2
?x?2
= e. Therefore
lim
x??

x + 1
x + 2
x
= lim
x??
"
1 +
1
?x ? 2
?x?2
# x
?x?2
= lim
x??
e
x
?x?2
= e
lim
x??
x
?x?2
= e
lim
x??
x?
(?x?2)?
= e?1
COMPARE: We have
lim
x??

x + 1
3x + 2
x
= 0
since
lim
x??
x + 1
3x + 2
=
1
3
and lim
x??

1
3
x
= 0
Similarly,
lim
x??

3x + 1
x + 2
x
= ? (D.N.E.)
since
lim
x??
3x + 1
x + 2
= 3 and lim
x??
3x = ? (D.N.E.)
19. Find lim
x??

1 +
1
x
x
.
14
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
19. Find lim
x??

1 +
1
x
x
.
Solution: We have
lim
x??

1 +
1
x
x
= e
by definition of e. Note, that we can’t use the approach described in Example 13, since it is based
of the formula (ln x)? =
1
x
. But this result was deduced from the fact that lim
x??

1 +
1
x
x
= e
(see Appendix A). This is Circular Reasoning which is a logical fallacy. However, one can apply
L’Hospital’s Rule to modifications of this limit.
20. Find lim
x?0+
(1 + sin 7x)cot 5x.
Solution: Note that lim
x?0+
(1 + sin 7x)cot 5x is 1? type of an indeterminate form. Put
y = (1 + sin 7x)cot 5x
then
ln y = ln
??
(1 + sin 7x)cot 5x

= cot 5x ln(1 + sin 7x) =
cos 5x ln(1 + sin 7x)
sin 5x
We have
lim
x?0+
cos 5x ln(1 + sin 7x)
sin 5x
= lim
x?0+
cos 5x · lim
x?0+
ln(1 + sin 7x)
sin 5x
= lim
x?0+
ln(1 + sin 7x)
sin 5x
=

0
0

= lim
x?0+
[ln(1 + sin 7x)]?
(sin 5x)? = lim
x?0+
1
1+sin 7x · (1 + sin 7x)?
cos 5x · (5x)?
= lim
x?0+
1
1+sin 7x · cos 7x · (7x)?
cos 5x · 5
= lim
x?0+
1
1+sin 7x · cos 7x · 7
cos 5x · 5
=
1
1+0 · 1 · 7
1 · 5
=
7
5
Therefore
lim
x?0+
(1 + sin 7x)cot 5x = e7/5
In short,
y = (1 + sin 7x)cot 5x =? ln y = cot 5x ln(1 + sin 7x) =
cos 5x ln(1 + sin 7x)
sin 5x
We have
lim
x?0+
cos 5x ln(1 + sin 7x)
sin 5x
= lim
x?0+
ln(1 + sin 7x)
sin 5x
= lim
x?0+
[ln(1 + sin 7x)]?
(sin 5x)?
= lim
x?0+
1
1+sin 7x · cos 7x · 7
cos 5x · 5
=
1
1+0 · 1 · 7
1 · 5
=
7
5
Therefore lim
x?0+
(1 + sin 7x)cot 5x = e7/5.
15
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
Appendix A
THEOREM: The function f(x) = sin x is differentiable and
f?(x) = cos x
Proof: We have
(sin x)? = lim
h?0
sin(x + h) ? sin x
h
[We use sin( + ) = sin  cos  + cos  sin ]
= lim
h?0
sin x cos h + cos x sin h ? sin x
h
= lim
h?0

sin x cos h ? sin x
h
+
cos x sin h
h

= lim
h?0

sin x(cos h ? 1)
h
+ cos x ·
sin h
h

= sin x lim
h?0
cos h ? 1
h
+ cos x lim
h?0
sin h
h
=sin x · 0 + cos x · 1 = cos x
THEOREM: The function f(x) = loga x is differentiable and
f?(x) =
1
x ln a
Proof: We have
d
dx
(loga x) = lim
h?0
loga(x + h) ? loga x
h
= lim
h?0

1
h
loga

x + h
x

= lim
h?0

1
h
loga

1 +
h
x

= lim
h?0

1
x ·
x
h
loga

1 +
h
x

=
1
x
lim
h?0
"
loga

1 +
h
x
x/h
#
=
1
x
lim
h?0
"
loga

1 +
h
x
1/(h/x)
#
=
1
x
loga
"
lim
h?0

1 +
h
x
1/(h/x)
#
=
h
lim
u?0
(1 + u)1/u = e
i
=
1
x
loga e =
1
x
ln e
ln a
=
1
x ln a
16
Section 3.7 Indeterminate Forms and L’Hospital’s Rule 2010 Kiryl Tsishchanka
Appendix B
1. Find lim
x?0
sin x
2x
.
Solution 1: We have
lim
x?0
sin x
2x
=

0
0

= lim
x?0
(sin x)?
(2x)? = lim
x?0
cos x
2
=
cos 0
2
=
1
2
Solution 2: We have
lim
x?0
sin x
2x
=

0
0

=
1
2
lim
x?0
sin x
x
=
1
2 · 1 =
1
2
2. Find lim
x?0
sin 2x
x
.
Solution 1: We have
lim
x?0
sin 2x
x
=

0
0

= lim
x?0
(sin 2x)?
x? = lim
x?0
2 cos 2x
1
=
2 cos 0
1
= 2
Solution 2: We have
lim
x?0
sin 2x
x
=

0
0

= 2 lim
x?0
sin 2x
2x
= 2 · 1 = 2
3. Find lim
x?0
sin 2x
5x
.
Solution 1: We have
lim
x?0
sin 2x
5x
=

0
0

= lim
x?0
(sin 2x)?
(5x)? = lim
x?0
2 cos 2x
5
=
2 cos 0
5
=
2
5
Solution 2: We have
lim
x?0
sin 2x
5x
=

0
0

=
2
5
lim
x?0
sin 2x
2x
=
2
5 · 1 =
2
5
4. Find lim
x?0
sin 2x
sin 5x
.
Solution 1: We have
lim
x?0
sin 2x
sin 5x
=

0
0

= lim
x?0
(sin 2x)?
(sin 5x)? = lim
x?0
2 cos 2x
5 cos 5x
=
2 cos 0
5 cos 0
=
2
5
Solution 2: We have
lim
x?0
sin 2x
sin 5x
=

0
0

= lim
x?0
sin 2x
2x · 2x
sin 5x
5x · 5x
= lim
x?0
1 · 2x
1 · 5x
= lim
x?0
2
5
=
2
5
17