historically, this result first appeared in l'hôpital's 1696 treatise, which was the...
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Intermediate Forms and L’Hospital Rule
(also L’Hopital and Bernoulli Rule)
Historically, this result first appeared in L'Hôpital's 1696 treatise, which was the first textbook on differential calculus.
Within the book, L'Hôpital thanks the Bernoulli brothers for their assistance and their discoveries.
An earlier letter by John Bernoulli gives both the rule and its proof, so it seems likely that Bernoulli discovered the rule.
Definition: Indeterminate Limit/Form
If exists, where , the limit is said to be indeterminate.
lim𝑥→𝑐
𝑓 (𝑥)𝑔(𝑥 )
The following expressions are indeterminate forms:
These expressions are called indeterminate because you cannot determine their exact value in the indeterminate form.
However, it is still possible to solve these in many cases due to L'Hôpital's rule.
We used a geometric argument to show that:
You have previously studied limits with the indeterminate form
I Indeterminate Form
Example:
Example: 42222
)2)(2(
2
4limlimlim
22
2
2
x
x
xx
x
x
xxx
Example:
xx
x
xxx
x
x
xxx 2sin
1
3cos
1
1
3sin
2sin3cos3sin
2sin
3tanlimlimlim
000
Example:
2
3)1)(1)(1(
2
3
2sin
2
3cos
1
3
3sin
2
3limlimlim
02003
x
x
xx
x
xxx
Some limits can be recognized as a derivative
h
afhafaf
h
)()()( lim
0
Recognizing a given limit as a derivative (!!!!!!)
Example:
Example:
Example:
Tricky, isn’t it? A lot of grey cells needed.
Not all forms are like those.
with the knowledge given to you
by now you get a doctorate at 17 and get to quit the school right now.
If you can find
L’Hospital rule for indeterminate form
Let f and g be real functions which are continuous on the closed interval [a, b] and differentiable on the open interval (a, b) . Suppose that . Then:
lim𝑥→𝑐
𝑓 (𝑥)𝑔(𝑥 )
=lim𝑥→𝑐
𝑓 ′ (𝑥)𝑔 ′ (𝑥)
provided that the second limit exists.
Example:
Example: 4)2(21
2
2
4limlim
2
2
2
x
x
x
xx
2
3
)1(2
)1(3
2cos2
3sec3
2sin
3tan 2
00limlim
x
x
x
x
xxExample:
12
1
)8(3
1
)8(3
1
1
)1()8(3
128
32
32
0
32
0
3
0limlimlim
h
h
h
h
hhhExample:
2
1
22
11limlimlim
002
0
x
x
x
x
x
x
e
x
e
x
xeExample:
2
33sin
1
sin
3
21cos
limlim33
x
x
x
xxExample:
01
01cos
2
11cos
2
1sin
1
limlimlim2
32
x
x
xx
x
x
xxxx
Example:
Example:
01
)0(2
cos
2
sin1sin
1
limlimlim0
2
0
2
y
y
y
y
x
xyyx xy 1
Example:
Example: lim𝑥→ 2
sin (𝑥2−4 )𝑥−2
=lim𝑥→ 2
2𝑥 cos (𝑥2−4 )1
=4
Suppose that instead of , we have that and as . Then:
Corollary for indeterminate form
lim𝑥→𝑏−
𝑓 (𝑥)𝑔 (𝑥)
= lim𝑥→𝑏−
𝑓 ′ (𝑥)𝑔 ′ (𝑥)
provided that the second limit exists
II Indeterminate Form
2
3
4
6
34
56
132
753limlimlim 2
2
xxx x
x
xx
xxExample:
Easier: 2
3
002
00313
2
753
132
753
132
753limlimlimlim
2
2
222
2
222
2
2
2
xxxx
xx
xx
xxx
xx
xxx
xx
xx
xx
=
4
18
4
9
12
43limlimlim
2
2
3 x
x
x
x
x
xxx
Example:
12
432
3
lim x
x
x
0
3
00
0312
43
3
3
limxx
xx
limit does not exist.
3
2
33
22
13
12
1
31
2
)1ln(
)1ln(24
4
22
3
3
2
2
3
2
limlimlimlim
xx
xx
xx
xx
x
xxx
x
x
xxxxExample:
02
0
222
1
1ln 22
0
3
03020limlimlimlim
x
x
x
x
x
x
x
xxxx
=
=
=Example:
Example:
14
4
1142
8
1
142
22
limlimlimx
xx
x
x
x
xxx
Rule does not help in this situation. It is a pure pain. In the situations like this one divide in you mind denominator and numerator with highest exponent of
Example:
2/11
/14
1
14limlim
2
x
x
x
x
xx
III Indeterminate Form
0 ∙ ∞ use algebra to convert the expression to a fraction (0 ), and then apply L'Hopital's Rule
Example: lim𝑥→𝜋 /2−
(𝑥− 𝜋2 ) tan𝑥= (0 ∙∞ )= lim𝑥→ 𝜋/2−
𝑥−𝜋2
cot𝑥=( 0
0 )= lim𝑥→𝜋 /2−
1
−𝑐𝑠𝑐2𝑥=−1
Example: 0)(1
1
1ln
ln limlimlimlimlim0
2
02000
x
x
x
x
x
x
xxx
xxxxx
x
xx
xxx
x
xxx
xxxx
tansin
cotcsc
1
csc
lnln)(sin limlimlimlim
0000
Example:
0)0)(1(tansin
limlim00
xx
x
xx
Example: 1
sin1
1sin1sin limlimlim
0
y
y
x
xxx
yxx xy 1
Example:
Example:
A limit problem that leads to one of the expressions
is called an indeterminate form of type
(+∞ )− (+∞ ) , (−∞ )− (−∞ ) , (+∞ )+(−∞ ) , (−∞ )+(+∞)
Such limits are indeterminate because the two terms exert conflicting influences on the expression; one pushes it in the positive direction and the other pushes it in the negative direction
IV Indeterminate Form
Indeterminate forms of the type can sometimes be evaluated by combining the terms and manipulating the result to produce an indeterminate form of type
00𝑜𝑟∞∞
Example: convert the expression into a fraction by rationalizing
xxx
x
xx
xx
xx xxx sincos
1cos
sin
sin
sin
11limlimlim
000Example:
02
0
coscossin
sinlim
0
xxxx
x
x
Example:
20
2
0
cos1lnln)cos1ln( limlim
x
xxx
xx
2
1ln
2
sinln
cos1ln limlim
02
0 x
x
x
x
xx
Example:
Example:
V Indeterminate Form
Several indeterminate forms arise from the limit
These indeterminate forms can sometimes be evaluated as follows:
)()( xgxfy
)(ln)()(lnln )( xfxgxfy xg
)(ln)(ln limlim xfxgyaxax
1.
2.
3.
Take of both sides
Find the limit of both sides
)(ln)(lim xfxgax
4. If = L
5.
Example: 1∞
ln y = ln[(1 + sin 4x)cot x] = cot x ln(1 + sin 4x)
FindExample: 00
l n 𝑦=𝑥 ln 𝑥lim
𝑥→0+¿ ln𝑦= lim𝑥→0 +¿𝑥 ln𝑥=(0 ∙−∞ )= lim
𝑥→ 0+ ¿ ln𝑥
1𝑥
= lim
𝑥→ 0+ ¿
1
𝑥
−1
𝑥2
= lim𝑥→ 0+¿ (− 𝑥 )=0
¿
¿ ¿
¿ ¿¿ ¿¿ ¿ ¿¿
¿
lim𝑥→ 0+¿𝑦= lim
𝑥→0+ ¿𝑒 ln𝑦=𝑒0=1
¿ ¿¿¿
xx = (eln x)x = ex ln x
lim𝑥→0+¿𝑥𝑥¿
¿
lim𝑥→0+¿𝑥𝑥=1¿
¿
xx
xe
2)1(lim
Example: Find 0
xxey
2)1(
x
eey
xxx )1ln(2
)1(lnln2
x
ey
x
xx
)1ln(2ln limlim
22
1
2
1
12
limlimlim
x
x
xx
x
x
x
x
x e
e
e
eee
xx
x
e2
)1(lim2e
(∞∞ )
(∞∞ )
FindExample: x
x
x1
0
coslim
1
xxy1
)(cos
x
xxy x )ln(cos)(coslnln
1
x
xy
xx
)ln(cosln limlim
00( 0
0 ) 0tanlim
0
xx
x
x
x1
0
coslim
10 e