Math Problem Statement

The graph of f is given. Use the graph fo compute the quantifies asked for. ( if an answer does not exist, enter DNE)

The xy-coordinate plane is given. • The curve with two parts enters the window in the second quadrant just above the x-axis, goes up and right getting more steep, passes through the point (-4, 4), exits the window in the second quadrant nearly vertically to the left of the y-axis, reenters in the first quadrant nearly vertically to the right of the y-axis, goes down and right getting less steep, crosses the x-axis at the open point (4, 0), and exits the window in the fourth quadrant nearly horizontally above y = -4. • There is a closed point at the origin. • There is a closed point at (4, 4)

(a) Lim f(x) x-> 0^-

(b) Lim f(x) x-> 4^+

(c) Lim f(x) x -> 0

(d) Lim f(x) x-> 4

(e) f(0)

(f) f(4)

Solution

Given the description of the graph, let's compute the requested limits and function values.

(a) limx0f(x)\lim_{{x \to 0^-}} f(x)

This limit represents the behavior of the function as xx approaches 0 from the left. According to the description, the curve in the second quadrant goes up and right steeply and exits the window to the left of the y-axis, re-entering the first quadrant near the origin. There is a closed point at the origin.

Since the graph approaches the origin from the left (from negative values of xx), and there is a closed point at (0, 0), we conclude that: limx0f(x)=0\lim_{{x \to 0^-}} f(x) = 0

(b) limx4+f(x)\lim_{{x \to 4^+}} f(x)

This limit represents the behavior of the function as xx approaches 4 from the right. The graph crosses the x-axis at (4,0)(4, 0) with an open point, but there is no further description of the function for values of x>4x > 4.

From the description, as xx approaches 4 from the right, it approaches an open point at (4,0)(4, 0), meaning the limit does exist and equals: limx4+f(x)=0\lim_{{x \to 4^+}} f(x) = 0

(c) limx0f(x)\lim_{{x \to 0}} f(x)

This limit considers the behavior of the function as xx approaches 0 from both sides. From part (a), the left-hand limit is 0. Since the right-hand limit (from the right of x=0x = 0) approaches the closed point at (0,0)(0, 0), both sides give the same value.

Thus, the two-sided limit is: limx0f(x)=0\lim_{{x \to 0}} f(x) = 0

(d) limx4f(x)\lim_{{x \to 4}} f(x)

This limit considers the behavior of the function as xx approaches 4 from both sides. From part (b), the limit from the right-hand side is 0. From the left-hand side, the curve passes through the point (4,4)(4, 4), indicating the left-hand limit is 4.

Since the left-hand and right-hand limits do not match, the two-sided limit does not exist: limx4f(x)=DNE\lim_{{x \to 4}} f(x) = \text{DNE}

(e) f(0)f(0)

From the description, there is a closed point at the origin (0,0)(0, 0), meaning the function value at x=0x = 0 is: f(0)=0f(0) = 0

(f) f(4)f(4)

There is a closed point at (4,4)(4, 4), indicating that the function value at x=4x = 4 is: f(4)=4f(4) = 4

Summary of Results:

(a) limx0f(x)=0\lim_{{x \to 0^-}} f(x) = 0
(b) limx4+f(x)=0\lim_{{x \to 4^+}} f(x) = 0
(c) limx0f(x)=0\lim_{{x \to 0}} f(x) = 0
(d) limx4f(x)=DNE\lim_{{x \to 4}} f(x) = \text{DNE}
(e) f(0)=0f(0) = 0
(f) f(4)=4f(4) = 4


Would you like more details on these computations? Here are 5 related questions to expand this topic:

  1. How do one-sided limits differ from two-sided limits?
  2. What are the conditions for a limit to exist at a given point?
  3. How do open and closed points on a graph affect the function's value?
  4. How can discontinuities like jumps and holes be identified using limits?
  5. What role do limits play in understanding the continuity of a function?

Tip: When evaluating limits graphically, always check the behavior of the function from both the left and right to identify possible discontinuities or mismatches.

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Math Problem Analysis

Mathematical Concepts

Limits
Continuity
Graphical Analysis

Formulas

Limit definitions for one-sided and two-sided limits

Theorems

Limit existence theorem
Continuity and discontinuity conditions

Suitable Grade Level

Grades 11-12