Lesson 11.2 Techniques for Evaluating Limits Essential Question: How do you evaluate limits that cannot be solved through use of direct substitution?

Before we start… Evaluate the limit. lim 𝑥→−3 𝑥 2 +𝑥−6 𝑥+3

What happens when you get 0/0? If you use direct substitution and get 0 0 , then you have no real answer because you have the indeterminate form. You must use a dividing out technique or rationalizing technique before you can determine that the limit does not exist.

Dividing Out Technique Factor the numerator Factor the denominator Cancel Use direct substitution to evaluate

Evaluate the limit. lim 𝑥→−3 𝑥 2 +𝑥−6 𝑥+3

Evaluate the limit. lim 𝑥→2 𝑥 2 +2𝑥−8 𝑥−2

Evaluate the limit. lim 𝑥→1 𝑥−1 𝑥 3 − 𝑥 2 +𝑥−1

Evaluate the limit. lim 𝑥→2 𝑥−2 𝑥 3 − 2𝑥 2 +2𝑥−4

Evaluate the limit. lim 𝑥→0 𝑥+2 2 −4 𝑥

Evaluate the limit. lim 𝑥→0 𝑥+3 2 −9 𝑥

Rationalizing Technique Multiply by the conjugate Simplify Use direct substitution to evaluate

Evaluate the limit. lim 𝑥→0 𝑥+1 −1 𝑥

Evaluate the limit. lim 𝑥→0 𝑥+9 −3 𝑥

Evaluate the limit. lim 𝑥→3 𝑥+1 −2 𝑥−3

Evaluate the limit. lim 𝑥→0 1 𝑥+4 − 1 4 𝑥

Evaluate the limit. lim 𝑥→0 1 𝑥−8 + 1 8 𝑥

For the function given by 𝑓 𝑥 = 𝑥 2 −1, find lim ℎ→0 𝑓 3+ℎ −𝑓 3 ℎ

For the function given by 𝑓 𝑥 = 2𝑥 2 +1, find lim ℎ→0 𝑓 2+ℎ −𝑓 2 ℎ

How do you evaluate limits that cannot be solved through use of direct substitution?

Ticket Out the Door Evaluate the limit. lim 𝑥→2 𝑥 2 −𝑥−2 𝑥−2