# Indeterminate Form (Part 2)

How to solve the limits of functions in indeterminate form [0/0, 0⋅∞]: examples and their solutions.

## Example 1: 0/0

If you put 1 into the *x*-s,

both of the numerator and the denominator become 0,

which yields 0/0 form.

To solve this,

find the factor that makes 0, (*x* - 1)

and cancel (*x* - 1).

To find (*x* - 1),

factor the numerator.

Factoring a quadratic trinomial

Cancel *x* - 1,

which makes 0/0 form.

Put 1 into the *x*.

Then (limit) = 3.

## Example 2: 0/0

To solve 0/0 form with a square root,

multiply the conjugate of the square root part

(√*x* + 7 + 3)

to both of the numerator and the denominator.

It's like rationalizing the numerator

to find (*x* - 2) factor,

which makes 0/0 form.

Rationalizing a denominator

Cancel *x* - 2,

which makes 0/0 form.

Put 2 into the *x*.

## Example 3: 0 × ∞

If you put 1 into the *x*-s,

1/*x* becomes ∞

and 6/(*x* + 3) - 2 becomes, 2 - 2, 0,

which yields 0 × ∞ form.

To solve this,

change 6/(*x* + 3) - 2

to find the factor (*x*). (blue)

Adding and subtracting rational expressions

Cancel *x*,

which makes 0 × ∞ form.

Put 0 into the *x*.