# 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

3n^{2}.

The highest order term in the denominator is

n^{2}.

The orders of 3n^{2} and n^{2} are the same: 2.

Then write the coefficients of the terms:

3/1.

This is the limit value.

3/1 = 3

So 3 is the answer.

## Example 2

### Example

As n → ∞,

the numerator goes to ∞

and the denominator goes to ∞.

So this is an indeterminate form ∞/∞.

### Solution

3^{n + 4}

= 3^{n}⋅3^{4}

= 81⋅3^{n}

+2^{2n + 1}

= +2⋅2^{2n}

= +2⋅4^{n}

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⋅4^{n}.

And the fastest term in the denominator is

+4^{n}.

The bases of +2⋅4^{n} and +4^{n}

are the same: 4.

Then write the coefficients of the terms:

2/1.

This is the limit value.

2/1 = 2

So 2 is the answer.

## 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

√4n^{2} ... (= 4n^{2/2}) and +√n^{2} ... (= +n^{2/2}).

Rational Exponent

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 = √2^{2} = 2

√1 = 1

Square Root

2 + 1 = 3

3/3 = 1

So 1 is the answer.

## Example 4

### Example

As n → ∞,

√n^{2} + 6n - 12 goes to ∞

and n goes to ∞.

So this is an indeterminate form ∞ - ∞.

### Solution

To solve ∞ - ∞ form,

multiply the conjugate of (√n^{2} + 6n - 12 - n),

(√n^{2} + 6n - 12 + n),

to both of the numerator and the denominator.

(√n^{2} + 6n - 12 - n)(√n^{2} + 6n - 12 + n)

= (n^{2} + 6n - 12 - n^{2})

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

Cancel n^{2} and -n^{2}.

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

√n^{2} ... (= n^{2/2}) 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

So 3 is the answer.

## Example 5

### Example

As n → ∞,

√n^{2} + n goes to ∞

and √n^{2} - n goes to ∞.

So the denominator is an indeterminate form ∞ - ∞.

### Solution

To solve ∞ - ∞ form,

multiply the conjugate of (√n^{2} + n - √n^{2} - n),

(√n^{2} + n + √n^{2} - n),

to both of the numerator and the denominator.

(√n^{2} + n - √n^{2} - n)(√n^{2} + n + √n^{2} - n)

= n^{2} + n - (n^{2} - n)

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

-(n^{2} - n) = -n^{2} + n

Cancel n^{2} and -n^{2}.

+n + n = 2n

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

√n^{2} ... (= n^{2/2}) and +√n^{2} ... (= +n^{2/2}).

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

So 7 is the answer.

## 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 x^{2} + x - 2.

2⋅(-1) = -2

2 - 1 = +1

So x^{2} + x - 2 = (x + 2)(x - 1).

Factor a Quadratic Trinomial

Cancel (x - 1) factors.

Find the limit value.

Put 1 into (x + 2).

Limit of a Function

1 + 2 = 3

So 3 is the answer.

## 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 = √3^{2} = 3

3 + 3 = 6

So 1/6 is the answer.

## 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).

Add and Subtract Rational Expresssions

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

So -2/3 is the answer.