Multiple Choice Questions
Calculus: Limits and Derivatives (Introductory)
Grade: 10
Topic: Calculus – Limits and Derivatives (Introductory)
Question 1:
What is the limit of the function f(x) = (3x^2 – 2x + 1)/(2x^2 + 5x – 3) as x approaches 2?
a) 1/2
b) 2/3
c) 3/4
d) 4/5
Answer: a) 1/2
Explanation: To find the limit, we substitute the value of x into the function. Plugging in x = 2, we get f(2) = (3(2)^2 – 2(2) + 1)/(2(2)^2 + 5(2) – 3) = 11/15. Therefore, the limit is 11/15, which is option a).
Example:
Let\’s consider the function f(x) = (x^2 + 3x – 2)/(2x – 1). To find the limit as x approaches 3, we substitute x = 3 into the function. Plugging in x = 3, we get f(3) = (3^2 + 3(3) – 2)/(2(3) – 1) = 22/5. Therefore, the limit is 22/5.
Question 2:
What is the limit of the function g(x) = (2x^3 – 4x^2 + 3x – 1)/(x^3 – 2x^2 + x – 1) as x approaches 1?
a) 1
b) 2
c) 3
d) 4
Answer: a) 1
Explanation: To find the limit, we substitute the value of x into the function. Plugging in x = 1, we get g(1) = (2(1)^3 – 4(1)^2 + 3(1) – 1)/(1^3 – 2(1)^2 + 1 – 1) = 0/0. This is an indeterminate form, so we need to simplify the function. By factoring both the numerator and denominator, we can cancel out the common factors. After simplification, we find that g(x) = 2 for all values of x except x = 1. Therefore, the limit is 2.
Example:
Consider the function g(x) = (x^3 – 5x^2 + 6x – 2)/(2x^3 – 3x^2 + 2x – 1). To find the limit as x approaches 2, we substitute x = 2 into the function. Plugging in x = 2, we get g(2) = (2^3 – 5(2)^2 + 6(2) – 2)/(2(2)^3 – 3(2)^2 + 2(2) – 1) = 0/0. Again, this is an indeterminate form, so we need to simplify the function. By factoring both the numerator and denominator, we can cancel out the common factors. After simplification, we find that g(x) = 1 for all values of x except x = 2. Therefore, the limit is 1.
(Note: The examples provided are simplified versions of the given questions in order to illustrate the concept. The actual questions may involve more complex functions and calculations.)