Solving for Limits of Functions (Algebra)

There are 3 main ways to evaluate the limits of functions using algebra:

1. Direct substitution
2. Factoring and cancelling terms (hint: p(x)/q(x))
3. Conjugating (hint: square roots)

Let’s examine examples of how to use each of these methods.

Direct Substitution

Example 1 – Find the limit of f(x) = functions 1 when x approaches x = -3

Substitute x= -3 inside the equation.

functions 2

Example 2 – Find the limit of g(x) = x^3 + 2x + 2 when x = 5.

Substitute x = 5 into the equation:

(5) ^3 + 2(5) + 2 = 125 + 10 + 2 = 137

Factoring and Cancelling Terms

Use this method when you have a rational function that contains a polynomial in the numerator, denominator, or both.

Example 1 – Find the limit of f(x) = functions 3 when x approaches x = 2.

Notice that by using direct substitution, we obtain 20/0, which is undefined. Thus, factoring is the next best method.

Let’s factor the numerator:

x^2 + -5x + 6 = (x-3)(x-2)

Notice that both the numerator and denominator have (x-2). We can cancel that term in the numerator and denominator. Then we are left with:

functions 4

Substitute x = 2.

2-3 = 1.

The limit of f(x) as x approaches x = 2 is -1.

Example 2 – Find the limit of f(x) = functions 5 as x approaches x = -1

Factor the numerator:

2x3 + 3x + 1 = (2x+1) (x+1)

Factor the denominator:

x2 – 2x – 3 = (x-3) (x+1)

The numerator and denominator both have an (x+1). Cancelling those terms, you are left with:

functions 6

Substitute x = -1

functions 7

The limit of f(x) as x approaches x = -1 is 1/4.

Multiplying by The Conjugate

Use this method when you have a rational function with a square root expression in either the numerator or denominator.

Example 1 – Find the limit of f(x) = functions 8 as x approaches x = 7

Notice that by using direct substitution, we obtain a zero in the denominator and numerator, which makes the expression indeterminant. Therefore, another method must be used.

Notice we have the expression functions 9 in the numerator. We must find the conjugate of this expression (essentially, turning the minus sign into a plus sign), then we must multiply the numerator and denominator by the conjugate.

functions 10

Cancel the (x-7) from the top and bottom:

functions 11

Substitute x = 7 into the expression.

functions 12

The limit of f(x) as x approaches x= 7 is ⅙.

Example 2 – Find the limit of f(x) = functions 13 as x approaches x = 4.

Notice that plugging in x = 4 into the equation gives us 0/0, which is the indeterminate form. Therefore, the conjugate method must be used.

Find the conjugate (change the sign of the operation) of the expression in the denominator.

functions 14

Multiply the numerator and denominator by the conjugate:

functions 15

The numerator and denominator contain (x-4). Cancel this term.

functions 16

Substitute x = 4 into the equation:

functions 17

The limit of f(x) as x approaches x=4 is 4.