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AP Calculus AB: 2.2.3 Two Techniques for Evaluating Limits

Mathematics7 CardsCreated 3 months ago

This flashcard set explains techniques for evaluating limits that initially yield indeterminate forms, focusing on simplifying compound fractions and using conjugates to eliminate radicals. It emphasizes how these algebraic methods help resolve 0/0 forms to find the correct limit values.

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Two Techniques for Evaluating Limits

  • When evaluating the limit of a compound fraction, try to simplify the fraction by finding the lowest common denominator.

  • An expression involving a binomial can often be simplified by multiplying by the conjugate of the binomial. Given a binomial expression ( a + b ), the conjugate is the expression ( a – b ).

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Key Terms

Term
Definition

Two Techniques for Evaluating Limits

  • When evaluating the limit of a compound fraction, try to simplify the fraction by finding the lowest common denominator.

  • An ...

note

  • Attempting direct substitution with this limit results in an indeterminate form.

  • This expression is a compound fraction; it ...

Evaluate the limit lim x→2 x−√5x−6/x^2−4.

−1/16

Evaluate the limit lim h→0 1/(1+h)^2−1/h.

-2

Evaluate the limit lim x→0 x/√x+4−2.

4

Evaluate the limit lim x→2 1/x−1/2 / x−2.

-1/4

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AP Calculus AB: 2.2.3 Two Techniques for Evaluating Limits
TermDefinition

Two Techniques for Evaluating Limits

  • When evaluating the limit of a compound fraction, try to simplify the fraction by finding the lowest common denominator.

  • An expression involving a binomial can often be simplified by multiplying by the conjugate of the binomial. Given a binomial expression ( a + b ), the conjugate is the expression ( a – b ).

note

  • Attempting direct substitution with this limit results in an indeterminate form.

  • This expression is a compound fraction; it has a fraction in the numerator. You will need to simplify the numerator by finding a common denominator.

  • Dividing by a fraction is accomplished by multiplying by its reciprocal. Cancellation removes the 0/0 culprit.

  • Now direct substitution produces the value of the limit.

  • Here is another limit where direct substitution results in an indeterminate form. However, there is nothing to factor and nothing to combine.

  • You can remove the radical from the numerator if you multiply by its conjugate. The numerator is of the form a – b, so multiply the numerator and denominator by a + b. In this way you are essentially multiplying by 1.

  • Notice that the radical is now in the denominator. You may not think that you have made any progress, but now you can cancel the factors of x in the numerator and denominator.

  • Direct substitution then produces the value of the limit.

Evaluate the limit lim x→2 x−√5x−6/x^2−4.

−1/16

Evaluate the limit lim h→0 1/(1+h)^2−1/h.

-2

Evaluate the limit lim x→0 x/√x+4−2.

4

Evaluate the limit lim x→2 1/x−1/2 / x−2.

-1/4

Evaluate the limit lim x→1 [1/x−1 − 2/x^2−1].

HINT: Remember to combine the two fractions using a common denominator before evaluating the limit.

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