Transcript click - Uplift Education

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
𝑓(𝑐)
𝑔(𝑐)
𝑓(𝑥)
lim
exists, where lim 𝑓 𝑐 = lim 𝑔 𝑐 = 0 , the limit 𝑥→𝑐
𝑔(𝑥)
𝑥→𝑐
𝑥→𝑐
is said to be indeterminate.
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.
0
I Indeterminate Form
0
0
You have previously studied limits with the indeterminate form
0
Example:
x2  4
( x  2)(x  2)

 lim x  2  2  2  4
Example: lim
lim
x 2 x  2
x 2
x2
x 2
Example:
Example:
We used a geometric
argument to show that:
lim
x 0
tan 3 x
 lim
sin 2 x
x 0

sin 3 x
sin 3 x
1
1
cos 3 x 


sin 2 x lim
1
cos 3 x sin 2 x
x 0
3
sin 3 x 
1 
2x  3
3
 lim
 lim
 lim
  (1)(1)(1) 
2  3 x0 3 x  x0 cos 3 x  2 x0 sin 2 x  2
2
Some limits can be recognized as a derivative
Recognizing a given limit as a derivative (!!!!!!)
f (a)  lim
h 0
3
Example:
Example:
Example:
lim
ℎ→0
f ( a  h)  f ( a )
h
8+ℎ−2
0
𝑑 3
=
=
𝑥
ℎ
0
𝑑𝑥
𝑥=8
1 −2/3
= 𝑥
3
𝑥=8
1
=
12
𝜋
𝜋
1
cos
+
ℎ
−
cos
cos 𝑥 − 2
0
3
3 = 𝑑 cos 𝑥
lim
=
=
lim
𝜋
ℎ→0
𝑥→𝜋 3
0
ℎ
𝑑𝑥
𝑥−3
3
= − sin 𝑥 𝜋 = −
2
𝑥= 3
sin 𝑥
0
sin 0 + 𝑥 − sin 0
𝑑
𝐹𝑖𝑛𝑑 lim
=
= lim
=
sin 𝑥
𝑥→0 𝑥
𝑥→0
0
𝑥
𝑑𝑥
= cos 𝑥
𝑥=0
=1
Tricky, isn’t it? A lot of grey cells needed.
𝑥=𝜋 3
𝑥=0
Not all forms
0
0
are like those.
If you can find
ln 𝑥
𝑥→1 𝑥−1
lim
ln 𝑥
𝑥→∞ 𝑥−1
𝑜𝑟 lim
with the knowledge given to you
by now you get a doctorate at 17 and get to quit the school right now.
L’Hospital rule for indeterminate form
0
0
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 ∃𝑐 ∈ (𝑎, 𝑏): 𝑓(𝑐) = 𝑔(𝑐) = 0. Then:
𝑓(𝑥)
𝑓′(𝑥)
lim
= lim
𝑥→𝑐 𝑔(𝑥)
𝑥→𝑐 𝑔′(𝑥)
provided that the second limit exists.
Example:
x2  4
2x

Example: lim
lim  2(2)  4
x 2 x  2
x 2 1
tan3x
3 sec 2 3x 3(1) 3
 lim


Example: lim
x 0 sin 2 x
x 0 2 cos 2 x
2(1) 2
3
Example: lim
h 0
Example:
1
2
(8  h) 3 (1)
8h 2
1
1
1
 lim 3
 lim


2
2
h
h 0
1
h 0
3(8  h) 3 3(8) 3 12
ex  x 1
ex 1
ex 1
 lim
 lim

lim
2
x 0
x 0
2x
x 0 2
2
x
cos x  1
Example: lim

x
3
1
2
2
3
x
x  0 0
 lim
 lim
1 2
x   cos 1
x   cos 1
1
x
x
x
x
2
x
lim
Example: x sin 1
 
Example:
 
 sin x
3
2



sin


lim 1
3
2
x 
x 
3
3
 

 
1
y2
2y
2( 0)
x2 


0
lim
lim
lim
1
x   sin
1
y  0 sin y
y 0 cos y
x
 
Example:
Example:
sin 𝑥 2 − 4
2𝑥 cos 𝑥 2 − 4
lim
= lim
=4
𝑥→2
𝑥→2
𝑥−2
1
y 1
x
∞
II Indeterminate Form
∞
Suppose that instead of 𝑓(𝑐) = 𝑔(𝑐) = 0,
−
we have that 𝑓(𝑥) → ∞ and 𝑔(𝑥) → ∞ as 𝑥 → 𝑏 . Then:
Corollary for indeterminate form
lim−
𝑥→𝑏
𝑓(𝑥)
𝑓′(𝑥)
= lim−
𝑔(𝑥) 𝑥→𝑏 𝑔′(𝑥)
3x 2  5 x  7

Example: lim 2
2
x

3
x

1
x  
Easier:
∞
∞
∞
∞
=
provided that the second limit exists
6x  5
6 3


lim
lim
4
x

3
4
2
x  
x  
3x 2 5 x 7
5 7


3

 2
2
2
2
2
3x  5 x  7
3 00 3
x
x
x 
x x 


2
lim
lim
lim
lim
2
3 1
3x 1
2
x   2 x  3 x  1
x   2 x
x  
x   2  0  0
2




x x2
x2 x2 x2
Example:
3x 3  4

lim
2
x  2 x  1
∞
∞
9x2
18 x


=lim
lim
4
x  4 x
x 
4
3x  4
x3  3  0  3 

1
2 x 2  1 lim
x  2
 3 00 0
x x
3
3
lim
x 
ln( x  1)

3
ln( x  1)
2
Example:
Example:
Example:
lim
x  
lim
x 0 
ln x

1 2
x
∞
∞
∞
∞
=
lim
x  
1
limit does not exist.
2x
2 x x3  1
2x4  2x 2
x2 1 
 lim 4

2
lim
2
2
2
3x
3
x   3 x x  1
x   3 x  3 x
x3  1




x3
x2
02
x 
 lim

0
lim
= lim
  2

  2

2
x

2
x 0
x 0
x 0
x3
8x
Example:
lim
x 
4x 2  1
2 4x 2  1
 lim
 lim
x 1
x 
1
x 
4x
4x  1
2

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 𝑥
lim
x  
4x2 1
4 1/ x
 lim
2
x 1
x   1  1 / x
III Indeterminate Form 0 ∙ ∞
0 ∙ ∞ use algebra to convert the expression to a fraction (0 ∙
1
0
=
0
0
),
and then apply L'Hopital's Rule
Example:
lim − 𝑥 −
𝑥→𝜋/2
𝜋
𝑥−2
𝜋
0
1
tan 𝑥 = 0 ∙ ∞ = lim −
=
= lim −
= −1
𝑥→𝜋/2 cot 𝑥
𝑥→𝜋/2 −𝑐𝑠𝑐 2 𝑥
2
0
Example:
1
ln x
 x2
x
lim x ln x  lim 1  lim  1  lim x  lim ( x)  0
x 0
x 0
x 0
x 0
x 0
2
x
x
1
ln x
 sin x tan x
x
Example: lim (sin x) ln x  lim
 lim
 lim


 csc x
  csc x cot x

x
x 0
x 0
x 0
x 0


 sin x 
  lim
 lim tan x   (1)(0)  0
x  x 0
 x 0 

Example:
Example:
Example:
 
x sin 1  lim
lim
x x
x 
 x 
sin 1
1
x
sin y
lim y  1
y 0
y 1
x
IV Indeterminate Form ∞ − ∞
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
Indeterminate forms of the type ∞ − ∞ can sometimes be evaluated by combining the
terms and manipulating the result to produce an indeterminate form of type
0
∞
𝑜𝑟
0
∞
Example:
convert the expression into a fraction by rationalizing
1 
cos x  1
1
 sin x  x 








lim
lim
Example: lim


sin x  x0  x sin x  x0 x cos x  sin x
x 0  x
 lim
x 0 
 sin x
0
 0
 x sin x  cos x  cos x 2
 1  cos x 
Example: lim ln(1  cos x)  lnx 2   lim ln
 
2
x 0
x 0  
x



 1  cos x 
 sin x 
1
 ln lim 

ln

ln



 

lim 2 x 
2
x


2
 x 0 
 x 0 
Example:
Example:
V Indeterminate Form 00 , ∞0 , 1∞
Several indeterminate forms arise from the limit
These indeterminate forms can sometimes be evaluated as follows:
1. y   f ( x)g ( x)
2. ln y  ln f ( x)g ( x)  g ( x) ln f ( x)
Take ln of both sides
3. lim ln y   lim g ( x) ln f ( x)
Find the limit of both sides
x a
x a
4. If lim g ( x) ln f ( x) = L
xa
5.
1∞
Example:
ln y = ln[(1 + sin 4x)cot x] = cot x ln(1 + sin 4x)
Example: Find
lim+ 𝑥 𝑥
𝑥→0
00
ln 𝑦 = 𝑥 ln 𝑥
xx = (eln x)x = ex ln x
1
ln 𝑥
lim+ ln 𝑦 = lim+ 𝑥 ln 𝑥 = (0 ∙ −∞) = lim+
= lim+ 𝑥 = lim+ −𝑥 = 0
1
1
𝑥→0
𝑥→0
𝑥→0
𝑥→0
𝑥→0
−
𝑥
𝑥2
lim+ 𝑦 = lim+ 𝑒 ln 𝑦 = 𝑒 0 = 1
𝑥→0
𝑥→0
lim+ 𝑥 𝑥 = 1
𝑥→0
(e  1)
lim
x
x
Example: Find
y  (e  1)
x
2
2
0
x

x
x

2
ln(
e
 1)

x
ln y  ln (e  1)



x
2
x
 2 ln(e x  1)
lim ln y  lim
x 
x 
x
lim
x  
 ex 

 2 x
 2e x
 e 1 

lim
x
1
e

1
x  
lim (e
x  
x
 1)
2
x
2
e

∞
∞
∞
 2e x
 lim
 2
x
∞
e
x  
Example: Find lim cos x
1
x
x0
y  (cos x)
1
x
1

1
ln(cos x)

x
ln y  ln (cos x)



x
0
ln(cosx)
 0
lim ln y  lim
x
x 0
x 0
lim  tan x   0
x 0 
lim cos x x  e 0  1
1
x0