Understanding Limits and Algebraic Techniques

Understanding Limits and Algebraic Techniques

Assessment

Interactive Video

Mathematics

10th - 12th Grade

Hard

Created by

Jackson Turner

FREE Resource

The video tutorial explains how to evaluate limits, even when a function is discontinuous at a point. It demonstrates using algebraic techniques to simplify functions, such as factoring the numerator using the difference of cubes. The tutorial shows how to simplify expressions by canceling common factors and then calculate the limit using direct substitution, ultimately finding the limit to be 27.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the issue encountered when trying to evaluate the limit at the point (5, 11)?

The numerator becomes zero.

The denominator becomes zero.

The function is undefined.

Both numerator and denominator become zero.

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why can't the limit be found using direct substitution initially?

The function is already simplified.

The function is not defined at the point.

The function is continuous.

The function is linear.

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What algebraic technique is suggested to find the limit?

Completing the square

Factoring the expression

Differentiation

Using the quadratic formula

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What formula is used to factor the numerator in this problem?

Difference of cubes

Difference of squares

Quadratic formula

Sum of cubes

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What are the values of 'a' and 'b' in the difference of cubes formula for this problem?

a = 5, b = 11

a = x, b = y

a = x - 8, b = y - 14

a = x + 8, b = y + 14

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What happens to the common factor in the numerator and denominator?

It is multiplied by itself.

It is divided by the denominator.

It is added to the numerator.

It is canceled out.

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the significance of finding a common factor in the numerator and denominator?

It increases the function's degree.

It changes the function's domain.

It allows for simplification of the function.

It helps in finding the derivative.

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