# Indeterminate Form

How to solve the limit in indeterminate form (∞/∞, ∞-∞, 0/0, 0*∞): 8 examples and their solutions.

## Example 1

### Example

As n → ∞,
the numerator goes to ∞
and the denominator goes to ∞.

So this is an indeterminate form ∞/∞.

### Solution

To solve ∞/∞ form,
find the fastest increasing terms
from the numerator and the denominator.

For polynomials,
the fastest increasing terms
are the highest order terms.

The highest order term in the numerator is
3n2.

The highest order term in the denominator is
n2.

The orders of 3n2 and n2 are the same: 2.

Then write the coefficients of the terms:
3/1.
This is the limit value.

3/1 = 3

## Example 2

### Example

As n → ∞,
the numerator goes to ∞
and the denominator goes to ∞.

So this is an indeterminate form ∞/∞.

### Solution

3n + 4
= 3n⋅34
= 81⋅3n

+22n + 1
= +2⋅22n
= +2⋅4n

Product of Powers

Power of a Power

To solve ∞/∞ form,
find the fastest increasing terms
from the numerator and the denominator.

An exponential term whose base is greater than 1
is faster than any polynomial term.
And the exponential term that has the bigger base
increases faster.

So the fastest term in the numerator is
+2⋅4n.

And the fastest term in the denominator is
+4n.

The bases of +2⋅4n and +4n
are the same: 4.

Then write the coefficients of the terms:
2/1.
This is the limit value.

2/1 = 2

## Example 3

### Example

As n → ∞,
the numerator goes to ∞
and the denominator goes to ∞.

So this is an indeterminate form ∞/∞.

### Solution

To solve ∞/∞ form,
find the fastest increasing terms
from the numerator and the denominator.

The highest order term in the numerator is
3n.

The highest order terms in the denominator are
4n2 ... (= 4n2/2) and +√n2 ... (= +n2/2).

Rational Exponents

The orders of the terms are the same: 1.

Then write the coefficients of the terms:
3/(√4 + √1).
This is the limit value.

4 = √22 = 2
1 = 1

Square Root

2 + 1 = 3

3/3 = 1

## Example 4

### Example

As n → ∞,
n2 + 6n - 12 goes to ∞
and n goes to ∞.

So this is an indeterminate form ∞ - ∞.

### Solution

To solve ∞ - ∞ form,
multiply the conjugate of (√n2 + 6n - 12 - n),
(√n2 + 6n - 12 + n),
to both of the numerator and the denominator.

(√n2 + 6n - 12 - n)(√n2 + 6n - 12 + n)
= (n2 + 6n - 12 - n2)

Product of a Sum and a Difference: (a + b)(a - b)

Cancel n2 and -n2.

As n → ∞,
the numerator goes to ∞
and the denominator goes to ∞.

So this is an indeterminate form ∞/∞.

So, to solve ∞/∞ form,
find the fastest increasing terms
from the numerator and the denominator.

The highest order term in the numerator is
6n.

The highest order terms in the denominator are
n2 ... (= n2/2) and +n.

The orders of the terms are the same: 1.

Then write the coefficients of the terms:
6/(√1 + 1).
This is the limit value.

1 + 1 = 1 + 1 = 2

6/2 = 3

## Example 5

### Example

As n → ∞,
n2 + n goes to ∞
and √n2 - n goes to ∞.

So the denominator is an indeterminate form ∞ - ∞.

### Solution

To solve ∞ - ∞ form,
multiply the conjugate of (√n2 + n - √n2 - n),
(√n2 + n + √n2 - n),
to both of the numerator and the denominator.

(√n2 + n - √n2 - n)(√n2 + n + √n2 - n)
= n2 + n - (n2 - n)

Product of a Sum and a Difference: (a + b)(a - b)

-(n2 - n) = -n2 + n

Cancel n2 and -n2.
+n + n = 2n

As n → ∞,
the numerator goes to ∞
and the denominator goes to ∞.

So this is an indeterminate form ∞/∞.

So, to solve ∞/∞ form,
find the fastest increasing terms
from the numerator and the denominator.

The highest order terms in the numerator are
n2 ... (= n2/2) and +√n2 ... (= +n2/2).

The highest order term in the denominator is
2n.

The orders of the terms are the same: 1.

Then write the coefficients of the terms:
7(√1 + √1)/2.
This is the limit value.

1 = 1

(1 + 1) = 2

7⋅2/2 = 7

## Example 6

### Example

As x → 1,
the numerator goes to 0
and the denominator goes to 0.

So this is an indeterminate form 0/0.

### Solution

To solve 0/0 form,
make a factor that makes the denominator 0, (x - 1),
and cancel the factors.

To make (x - 1),
factor x2 + x - 2.

2⋅(-1) = -2
2 - 1 = +1

So x2 + x - 2 = (x + 2)(x - 1).

Cancel (x - 1) factors.

Find the limit value.

Put 1 into (x + 2).

Limit of a Function

1 + 2 = 3

## Example 7

### Example

As x → 2,
the numerator goes to 0
and the denominator goes to 0.

So this is an indeterminate form 0/0.

### Solution

To solve 0/0 form,
make a factor that makes the denominator 0, (x - 2),
and cancel the factors.

To make (x - 2),
multiply the conjugate of (√x + 7 - 3),
(√x + 7 + 3),
to both of the numerator and the denominator.

(√x + 7 - 3)(√x + 7 + 3)
= x + 7 - 9

Product of a Sum and a Difference: (a + b)(a - b)

+7 - 9 = -2

Cancel (x - 2) factors.

Find the limit value.

Put 2
into 1/(√x + 7 + 3).

2 + 7 = 9

9 = √32 = 3

3 + 3 = 6

## Example 8

### Example

As x → 0,
[1/x] goes to ∞
and [6/(x + 3) - 2] goes to 0.

So this is an indeterminate form 0⋅∞.

### Solution

To solve 0⋅∞ form,
make a factor that makes the denominator 0, x,
and cancel the factors.

To make x,
simplify [6/(x + 3) - 2].

-2 = -2⋅(x + 3)/(x + 3).

6 - 2(x + 3) = 6 - 2x - 6

Cancel 6 and -6.

Cancel x factors.

Find the limit value.

Put 0
into -2/(x + 3).

0 + 3 = 3