Trigonometric Functions and Limits

Trigonometric Functions and Limits

Assessment

Interactive Video

Mathematics

11th Grade - University

Hard

Created by

Ethan Morris

FREE Resource

The video tutorial covers three challenging limit problems inspired by a student's calculus course. The instructor explains each problem in detail, starting with a polynomial limit problem solved by factoring. The second problem involves a limit with a square root, tackled using the conjugate method. The final problem requires knowledge of trigonometric identities to simplify and solve the limit. Throughout the video, the instructor emphasizes the importance of algebraic manipulation and understanding mathematical properties to solve complex limit problems effectively.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the first step in solving a limit problem involving a quadratic expression?

Apply L'Hôpital's Rule

Factor the expression

Substitute the limit value directly

Use the conjugate

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

When factoring the expression z^2 + 2z - 8, what are the factors?

(z + 5)(z - 1)

(z + 3)(z - 3)

(z + 2)(z - 4)

(z + 4)(z - 2)

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the purpose of multiplying by the conjugate in limit problems involving square roots?

To eliminate the square root

To simplify the expression

To factor the expression

To apply L'Hôpital's Rule

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What happens to the terms 1/x and 1/x^2 as x approaches infinity?

They remain constant

They approach 0

They approach 1

They approach infinity

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the limit of the expression as x approaches infinity in the second problem?

2

Infinity

0

1

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the cotangent of 2x in terms of sine and cosine?

1/cos(2x)

1/sin(2x)

sin(2x)/cos(2x)

cos(2x)/sin(2x)

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the double angle formula for sine used in the third problem?

sin(2x) = 1 - cos(2x)

sin(2x) = 2sin(x)cos(x)

sin(2x) = 2cos^2(x) - 1

sin(2x) = sin^2(x) - cos^2(x)

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