###### Calculus I

Exercices

Applications

Marginal analysis

# Calculus I: Limits Undetermined forms

### 1. The 7 undetermined forms

0/0, ∞/∞, ∞ - ∞, 0 x ∞, 1 , 00, ∞0.

When we get one of the above form, we have
to transform the expression of f(x) to evaluate
its limit.

### 2. The form 0/0

###### 2.1 Example:

lim (x2 - 25)/(x + 5)) = 0/0
x → - 5

Transforming f(x) gives:
(x2 - 25)/(x + 5)) = x - 5

Therefore:
 lim (x2 - 25)/(x + 5) x → - 5 = lim (x - 5) = - 10 x → - 5

###### 2.2 General rule:

When

lim f(x) = 0/0
x → a

We have necessarily

f(x) = (x - a)g1(x)/(x - a)g2(x) = g1(x)/g2(x)

Therefore:

 lim f(x) = x → a lim g1(x)/ g2(x) which is ≠ 0/0 x → a

### 3. The form ∞/∞

###### 3.1 Example:

lim (3x2 + 1)/(x2 + 5) = ∞/∞
x → + ∞

Transforming f(x) gives:
(3x2 + 1)/(x2 + 5) = (3x2 + 1)/x2/(x2 + 5)x2 = (3 + 1/x2)/(1 + 5/x2)

Therefore:
 lim (3x2 + 1)/(x2 + 5) = x → + ∞ lim (3 + 1/x2)/(1 + 5/x2) = 3/1 = 3 x → + ∞

###### 3.2. General rule:

When

lim f(x) = ±∞ /±∞
x → ±∞

We have necessarily a polynomial function of the form :
f(x) = g1(x)/g2(x).

We divide the numerator and denominator by the variable with the largest exponent.

Therefore:
 lim f(x) = x → ±∞ lim g1(x)/g2(x) x → ±∞

which is ≠ ∞/∞

###### 3.3 Other example:

f(x) = (x + 3 )/(x2 - 1)

lim f(x) = - ∞/+∞
x → - ∞
= lim (x + 3)/x2/(x2 - 1)/x2
x → - ∞
= lim (1/x + 3/x2)/(1 - 1/x2)
x → - ∞
= (0 + 0)/(1 - 0) = 0

### 4. The form ∞ - ∞

###### 4.1 Example 1 :

f(x) = x2 - x
x → ∞

lim f(x) = ∞ - ∞ : undetermined.
x → ∞

We factor using the term of the greatest degree:

x2 - x = x2(1 - 1/x)

Therefore:

lim f(x) = ∞( 1 - 0) = ∞
x → ∞

The rule is:
For a polynomial function, we take just the term of the greatest degree.

###### 4.2 Example 2:

f(x) = 3 x4 - x2 + 4
x → - ∞

lim f(x) = ∞ - ∞ : undetermined.
x → ∞

We use just the term of the greatest degree:

Therefore:

lim f(x) = lim 3 x4 = + ∞
x → ∞

###### 4.3 Example 3:

f(x) = √x - x
x → +∞

lim f(x) = ∞ - ∞ : undetermined.
x → + ∞

We use the conjugate quantity of f(x):

Therefore:

f(x) = √x - x = (√x - x)((√x + x)/(√x + x) =
(x - x2)/(√x + x) = x (1 - x)/(√x + x) =

Dividing the numerator and denominator by x gives:

f(x) = (1 - x)/((√x)/x + 1)

We have (√x)/x = √x/x2 = √(1/x)

Therefore:

f(x) = (1 - x)/(√(1/x) + 1)

So

lim f(x) = - √/(0 + 1) = - ∞
x → ∞

### 5. The form 0 x ∞

###### 5.1 Example 1 :

f(x) = x3 . 2/(x2 - 1) = 0 x ∞ : undetermined.
x → 0

Rule:
We use one of the above transformation.

x3 . 2/(x2 - 1) = 2 x3/(x2 - 1)

We divide the numerator and denominator by x3, that gives:

f(x) = 2 /(x - 1/x3)

Therefore:

lim f(x) = 2/(0 - ∞ ) = 0
x → 0

The rule is:
Use one of the above transformation.

###### 4.2 Example 2:

f(x) = |x - 2|/(x - 2)
x → 2;

lim f(x) = 0 x ∞ : undetermined.
x → 2

We have:
|x - 2| =
x - 2 if x > 2
- (x - 2) if x < 2

Therefore:

lim f(x) = (x - 2)/(x - 2) = 1
x → 2+
And
lim f(x) = - (x - 2)/(x - 2) = - 1
x → 2-
So
lim f(x)
x → 2

### 3. Exercises

1) 0/0

a)
lim (x2 - 6x + 8)/(2x2 - 3x - 2)
x → 2

b)
lim (2 - (x - 3)1/2)/(x2 - 49)
x → 7

c)
lim (x4 - 1)/((3 - x)1/2 + 2x)
x → - 1

d)
lim (3 - (x + 1)1/2)/(√x - 2√2)
x → + 8

e)
lim (3 + x )/( 1 + (3/x))
x → - 3

2) ∞/∞

a)
lim (3 + x )/(1 + x)
x → + ∞

b)
lim (4x3 - 2 x + 7)/(3x3 + 1)
x → + ∞

c)
lim (x2 + 7)/(x3 + 4)
x → - ∞

d)
lim (x + 1)/(2x3 - √(x2 - 2))
x → - ∞

e)
lim (x - 1)/(√(2x) + √(x)
x → + ∞

f)
lim (x2 + 2x)/(x + √(x2 - 1)
x → - ∞

3) ∞ - ∞

a)
lim (2x4 - x + 5)
x → + ∞

b)
lim (√(x2 + x) - x)
x → + ∞

c)
lim (x + √(2x2 + x - 1))
x → - ∞

d)
lim (1/x3 - 1/x)
x → 0

4) 0 x ∞

a)
lim (x(1/x - 1/(x - 1)))
x → + ∞

b)
lim ((1/(x - 3)3) (1/x - 1/3)) )
x → + 3

c)
lim (x √(1 + 1/x))
x → 0+;

d)
lim x/√(x)
x → + ∞

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