Multiple Choice
Question 1
What does the notation lim_{x→a} f(x) = L represent?
A
The value of f(x) at x = a is L.
B
The value of f(x) approaches L as x approaches a.
C
The value of f(x) is undefined at x = a.
D
The function f(x) has a vertical asymptote at x = a.
Solution: The notation lim_{x→a} f(x) = L means that f(x) approaches L as x approaches a, regardless of the value of f(a).
Question 2
Given a function f(x) with a vertical asymptote at x = 3, what is lim_{x→3+} f(x)?
A
0
B
∞
C
-∞
D
Does not exist
Solution: A vertical asymptote at x = 3 with the graph approaching ∞ from the right implies lim_{x→3+} f(x) = ∞.
Question 3
What is lim_{x→1} (3x^2 + 2x - 5)?
A
0
B
1
C
2
D
3
Solution: Using direct substitution, lim_{x→1} (3x^2 + 2x - 5) = 3(1)^2 + 2(1) - 5 = 0.
Question 4
Which type of discontinuity does f(x) = (x^2 - 1) / (x - 1) have at x = 1?
A
Removable discontinuity
B
Jump discontinuity
C
Infinite discontinuity
D
No discontinuity
Solution: The function simplifies to f(x) = x + 1 for x ≠ 1, so the discontinuity at x = 1 is removable.
Question 5
Given that lim_{x→c} f(x) = 2 and lim_{x→c} g(x) = 5, find lim_{x→c} [3f(x) - g(x)].
A
1
B
-1
C
6
D
11
Solution: Using limit laws, the limit is 3 * lim f(x) - lim g(x) = 3(2) - 5 = 6 - 5 = 1.
Question 6
If f(x) = 3x^2 - 5x + 2, what is lim_{x→2} f(x)?
A
0
B
4
C
12
D
20
Solution: Use direct substitution for the polynomial: f(2) = 3(4) - 5(2) + 2 = 12 - 10 + 2 = 4.
Question 7
For the function f(x), what is lim_{x→-2} f(x) if the function has a hole at x = -2 and approaches 4 from both sides?
A
-2
B
0
C
4
D
Does not exist
Solution: The limit as x approaches -2 is the value the function approaches, which is 4, regardless of the hole at x = -2.
Question 8
What is lim_{x→2} (5x - 7)?
A
3
B
5
C
7
D
10
Solution: Using direct substitution, lim_{x→2} (5x - 7) = 5(2) - 7 = 10 - 7 = 3.
Question 9
Which of the following functions is continuous at x = 0?
A
f(x) = |x|
B
f(x) = 1 / x
C
f(x) = (x^2 - 1) / (x - 1)
D
f(x) = sin(1 / x)
Solution: The function f(x) = |x| is defined at x = 0, lim_{x→0} |x| = 0, and f(0) = 0, so it is continuous at x = 0.
Question 10
What is the value of lim_{x→0^+} |x| / x?
A
-1
B
0
C
1
D
Does not exist
Solution: For x > 0, |x| = x, so the expression simplifies to x / x = 1. Therefore, the right-hand limit as x approaches 0 is 1.
Question 11
If f is continuous on [1, 5] with f(1) = -3 and f(5) = 5, which theorem guarantees a value c in (1, 5) such that f(c) = 0?
A
Extreme Value Theorem
B
Intermediate Value Theorem
C
Mean Value Theorem
D
Squeeze Theorem
Solution: The Intermediate Value Theorem states that if a function is continuous on a closed interval, it must take on every value between its values at the endpoints. Since 0 is between -3 and 5, such a c is guaranteed.
Question 12
If lim_{x→c} f(x) = 4 and lim_{x→c} g(x) = -2, evaluate lim_{x→c} [ (f(x) * g(x)) / (f(x) + g(x)) ].
A
-4
B
4
C
-2
D
2
Solution: Using algebraic properties of limits: numerator is 4 * (-2) = -8, and denominator is 4 + (-2) = 2. Therefore, -8 / 2 = -4.
Question 13
Evaluate lim_{x→-∞} (4x^3 - 2x) / (2x^3 + 5).
A
-2
B
0
C
2
D
Infinity
Solution: For rational functions at negative infinity where the degrees of the polynomial match, the limit is the ratio of the leading coefficients: 4 / 2 = 2.
Question 14
A particle's position is given by s(t) = t^3 - 2t^2. What is the average velocity over the interval [1, 3]?
A
5
B
8
C
10
D
16
Solution: Average velocity = (s(3) - s(1))/(3 - 1). s(3) = 27 - 18 = 9 and s(1) = 1 - 2 = -1, so (9 - (-1))/2 = 10/2 = 5.
Question 15
Which of the following is the correct interpretation of lim_{x→c} f(x) = L?
A
f(c) = L
B
f(x) approaches L as x approaches c, but f(c) may or may not equal L
C
f(x) = L for all x near c
D
f(x) is undefined at x = c
Solution: The limit lim_{x→c} f(x) = L means that f(x) approaches L as x approaches c, regardless of the value of f(c).
Question 16
For a function f(x) with a vertical asymptote at x = 2, what is lim_{x→2^-} f(x) if f approaches -∞ from the left?
A
∞
B
-∞
C
0
D
Does not exist
Solution: If the graph approaches -∞ as x approaches 2 from the left, then lim_{x→2^-} f(x) = -∞.
Question 17
What is lim_{x→-1} (2x^2 + 3x - 5)?
A
-6
B
-4
C
0
D
6
Solution: Polynomials are continuous, so substitute directly: 2(-1)^2 + 3(-1) - 5 = 2 - 3 - 5 = -6.
Question 18
Which type of discontinuity does f(x) = (x^2 - 4) / (x + 2) have at x = -2?
A
Removable discontinuity
B
Jump discontinuity
C
Infinite discontinuity
D
No discontinuity
Solution: The function simplifies to f(x) = x - 2 for x ≠ -2, so the discontinuity at x = -2 is removable.
Question 19
The position of a particle moving along a straight line is s(t) = t^3 - 6t^2 + 9t. What is the average velocity from t = 0 to t = 3?
A
0
B
3
C
9
D
-9
Solution: Average velocity is (s(3) - s(0)) / (3 - 0). Since s(0) = 0 and s(3) = 27 - 54 + 27 = 0, the average velocity is (0 - 0) / 3 = 0.
Question 20
Which of the following limits does NOT exist?
A
lim_{x→2} (3x^2 - 5)
B
lim_{x→0} (sin x)/x
C
lim_{x→0^+} 1/x
D
lim_{x→3} (x^2 - 9)/(x - 3)
Solution: As x approaches 0 from the right, 1/x grows without bound toward positive infinity, meaning the limit does not exist. All other options yield finite, real numbers.