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AP Calculus AB: 2.2.2 Limits and Indeterminate Forms

Mathematics12 CardsCreated 3 months ago

This set of flashcards introduces the fundamentals of evaluating limits, including direct substitution, identifying indeterminate forms, and applying algebraic techniques like factoring. It also covers special cases using the Squeeze Theorem and limit behavior near undefined points.

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Limits and Indeterminate Forms

  • To evaluate a limit at a value where a function is well behaved, substitute the value into the function expression.

  • Limits that produce indeterminate forms may or may not exist. An indeterminate form is a signal that more work is needed to evaluate the limit.

  • If direct substitution produces zero divided by a non-zero number, then the limit equals zero. If it produces a non-zero number divided by zero, the limit is undefined.

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Key Terms

Term
Definition

Limits and Indeterminate Forms

  • To evaluate a limit at a value where a function is well behaved, substitute the value into the function expression.

  • Limits t...

note

  • Always try direct substitution as your first step when evaluating a limit.

  • In this example direct substitution leads to the ...

Evaluate the limit lim x→−2 1/(2+x)^3.

The limit does not exist

Evaluate lim x→2 x+1/x^2−x−2.

The limit does not exist

Given g(x)= b−|x−a|,
h(x )= b+|x−a|,
and g(x) ≤ f(x) ≤h(x),
find lim x→a f(x).

b

Given g(x)= 2−x^2,
h(x)=2+x^2,
and g(x)≤f(x)≤h(x),
find lim x→0 f(x).

2

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AP Calculus AB: 2.2.2 Limits and Indeterminate Forms
TermDefinition

Limits and Indeterminate Forms

  • To evaluate a limit at a value where a function is well behaved, substitute the value into the function expression.

  • Limits that produce indeterminate forms may or may not exist. An indeterminate form is a signal that more work is needed to evaluate the limit.

  • If direct substitution produces zero divided by a non-zero number, then the limit equals zero. If it produces a non-zero number divided by zero, the limit is undefined.

note

  • Always try direct substitution as your first step when evaluating a limit.

  • In this example direct substitution leads to the value of limit 2.

  • When you first use direct substitution with this limit, it will produce the indeterminate form0/0. Try factoring the numerator and denominator. When you do, you will find that they have a common factor of x– 5. Cancel the common factors, keeping in mind that this is a limit.

  • The resulting expression can be evaluated using direct substitution.

  • Before you begin factoring, make sure you really have an indeterminate form. Zero divided by any number other than zero is not indeterminate. It equals zero.

  • In addition, any non-zero number divided by zero is not indeterminate. It is undefined. If a limit produces this type of expression, then the limit does not exist.

  • Here is a summary of quotients involving zero. Only the first one is an indeterminate form.

Evaluate the limit lim x→−2 1/(2+x)^3.

The limit does not exist

Evaluate lim x→2 x+1/x^2−x−2.

The limit does not exist

Given g(x)= b−|x−a|,
h(x )= b+|x−a|,
and g(x) ≤ f(x) ≤h(x),
find lim x→a f(x).

b

Given g(x)= 2−x^2,
h(x)=2+x^2,
and g(x)≤f(x)≤h(x),
find lim x→0 f(x).

2

Suppose that you are evaluating a limit, and after some simplification you reach the expression a0, where a≠0. What is the value of the limit?

The limit does not exist

Evaluate the limit lim x→−1 2/(x+1)^3.

The limit does not exist

Evaluate lim x→−5 x+5/x^2+7x+10.

−1/3

Evaluate lim x→3/2 8x^3−27/2x−3.

27

Evaluate lim x→−1/2 8x^2−2x−3/6x^2+x−1.

2

Evaluate limx→−3 x^2+5x+6/x+3.

-1