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WorksheetsLimits (infinite limits, limits at infinity & continuity)
Total questions: 22
Worksheet time: 2hrs 50mins
Evaluate the limit.
∞
−∞
11
DNE
Evaluate the limit.
∞
−∞
0
DNE
Evaluate the limit.
∞
−∞
0
DNE
What is the limit as x approaches 2?
0
4
41
DNE
What is the limit as -2 from the left?
∞
−∞
0
−41
Identify the vertical asymptote(s) in the function above. (Hint: The last three problems used this same function.)
x=2
x=±2
x=−2
There are no vertical asymptotes.
What is the limit of the function?
∞
−∞
0
DNE
What is the limit of the function?
∞
−∞
0
DNE
What is the limit of the function?
∞
−∞
0
DNE
∞
−∞
0
DNE
Evaluate the limit.
∞
−∞
0
5
Evaluate the limit.
∞
−∞
0
5
Identify the horizontal asymptote(s). (Hint: The last two question used the same function.)
y=5
y=−5
y=±5
There are no horizontal asymptotes.
Identify the discontinuities in the function.
x=1, 2, 3, 4, 5, 6
x=2, 3, 4, 5, 6
x=2, 3, 4, 5
x=2, 3, 4
Which discontinuities violate the first condition? Condition 1: f(a) is defined (a is in the domain of f).
x=3
x=2, 3
x=2, 3, 4
x=2, 4
Which discontinuities violate the second condition? (Condition 2: The limit as x approaches a exists.)
x=2, 3
x=2, 3, 4
x=2, 4
x=3, 4
Which discontinuities violate the third condition? (Condition 3: The value of f equals the limit of f at a.)
x=2, 3, 4
x=2, 3
x=3, 4
x=2, 4
For which values is the function below continuous?
f(x)=2x3+x+2
It is continuous on [−1,0], but not for all x.
It is continuous for some x, but not on [−1,0].
It is continuous for all x.
It is not continuous on any interval.
f(x)=2x3+x+2; (−1, 0)
What is the value at the left endpoint?
2
-1
0
1
f(x)=2x3+x+2; (−1, 0)
What is the value at the right endpoint?
-1
0
1
2
Can the Intermediate Value Theorem be used to show that f(x) has a solution on (-1, 0)? (Hint: The last two questions can be used to help answer this.)
f(x)=2x3+x+2
It can be used because the function is defined on (−1,0) and 0<f(−1)<f(0).
It can be used because the function is continuous on [−1,0] and the function is defined at x=−1 and x=0.
It can be used because the function is defined on (−1,0) and f(−1)<f(0)<0.
It can be used because the function is continuous on [−1,0] and 0 is between f(−1) and f(0).
f(x)=2x3+x+2
There is/are a solution(s) to the equation in (−1,0) at
x≈2.235
x≈−0.835
x≈−0.835, 2.235
There is not a solution in (-1, 0)
