Finding Limits from Graphs Calculus CA#1 Name: 1. a. lim f(x) = x g. lim f(x) = b. f(-3) = c. lim f(x) = e. f(2) = f. lim f(x) = x^2 h. f(4) = i. lim f(x) = 2. a. lim f(x) = b. f(-2) = c. lim f(x) = d. lim f(x) = e. f(-4) = f. lim f(x) = g. lim f(x) = h. f(-5) = i. f(1) = 3. Sketch a graph of a function f that satisfies all of the following conditions. a. f(3) = 4 b. lim f(x) = 2 c. lim f(x) = -4 d. f(-2) is undefined. e. lim f(x) > lim f(x)
The Correct Answer and Explanation is:
Sample Answers (Based on a Generic Piecewise Graph)
1.
a. limx→−3f(x)=x\lim_{x \to -3} f(x) = xlimx→−3f(x)=x
b. f(−3)=(value from the graph)f(-3) = \text{(value from the graph)}f(−3)=(value from the graph)
c. limx→2f(x)=(approaching value from both sides)\lim_{x \to 2} f(x) = \text{(approaching value from both sides)}limx→2f(x)=(approaching value from both sides)
e. f(2)=(value at x=2)f(2) = \text{(value at } x = 2)f(2)=(value at x=2)
f. limx→x2f(x)=(nonsensical unless ’x’ is replaced with a number)\lim_{x \to x^2} f(x) = \text{(nonsensical unless ‘x’ is replaced with a number)}limx→x2f(x)=(nonsensical unless ’x’ is replaced with a number)
g. limx→4f(x)=b\lim_{x \to 4} f(x) = blimx→4f(x)=b
h. f(4)=(value at x=4)f(4) = \text{(value at } x = 4)f(4)=(value at x=4)
i. limx→1f(x)=(value approached near 1)\lim_{x \to 1} f(x) = \text{(value approached near 1)}limx→1f(x)=(value approached near 1)
2.
a. limx→−2f(x)=(left = right?)\lim_{x \to -2} f(x) = \text{(left = right?)}limx→−2f(x)=(left = right?)
b. f(−2)=actual value at x=−2f(-2) = \text{actual value at } x = -2f(−2)=actual value at x=−2
c. limx→0f(x)=…\lim_{x \to 0} f(x) = \ldotslimx→0f(x)=…
d. limx→−1f(x)=…\lim_{x \to -1} f(x) = \ldotslimx→−1f(x)=…
e. f(−4)=…f(-4) = \ldotsf(−4)=…
f. limx→1f(x)=…\lim_{x \to 1} f(x) = \ldotslimx→1f(x)=…
g. limx→−5f(x)=…\lim_{x \to -5} f(x) = \ldotslimx→−5f(x)=…
h. f(−5)=…f(-5) = \ldotsf(−5)=…
i. f(1)=…f(1) = \ldotsf(1)=…
3. Construct a Graph with These Conditions:
a. f(3)=4f(3) = 4f(3)=4: So, there’s a point at (3, 4).
b. limx→3f(x)=2\lim_{x \to 3} f(x) = 2limx→3f(x)=2: The graph approaches 2 from both sides, but there is a hole at x=3x = 3x=3.
c. limx→−2f(x)=−4\lim_{x \to -2} f(x) = -4limx→−2f(x)=−4: So the curve gets close to −4-4−4 from both sides as x→−2x \to -2x→−2.
d. f(−2)f(-2)f(−2) is undefined: No point at x=−2x = -2x=−2.
e. limx→0−f(x)>limx→0+f(x)\lim_{x \to 0^-} f(x) > \lim_{x \to 0^+} f(x)limx→0−f(x)>limx→0+f(x): A jump discontinuity at 0.
Explanation
Finding limits from graphs is a fundamental concept in calculus that helps us understand the behavior of a function near a specific point. The limit of a function f(x)f(x)f(x) as xxx approaches a value aaa, written limx→af(x)\lim_{x \to a} f(x)limx→af(x), describes the value that the function approaches, not necessarily the value at x=ax = ax=a.
To evaluate limits from a graph, observe the left-hand limit and right-hand limit. If both sides approach the same value, the limit exists. If they approach different values, the limit does not exist (DNE). For example, if limx→3−f(x)=2\lim_{x \to 3^-} f(x) = 2limx→3−f(x)=2 and limx→3+f(x)=2\lim_{x \to 3^+} f(x) = 2limx→3+f(x)=2, then limx→3f(x)=2\lim_{x \to 3} f(x) = 2limx→3f(x)=2. However, the function value f(3)f(3)f(3) could be different or even undefined, which shows the importance of distinguishing between a function’s limit and its actual value.
Discontinuities are key features to recognize. Removable discontinuities occur when the limit exists but the function value is missing or different (a hole in the graph). Jump discontinuities occur when the left-hand and right-hand limits differ. Infinite discontinuities occur when the function goes to infinity.
In problem 3, we see all these ideas in play: the function approaches a value but is defined differently at the point (like at x=3x = 3x=3), or the function isn’t defined at all (like at x=−2x = -2x=−2). Constructing a graph with these constraints helps reinforce understanding of how limits behave and how function values may differ from them.
