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One-sided limits and discontinuities

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55144912
The graph represents one function \(f\) with a jump at \(x=1\). The open circles show the heights approached by the graph from the two sides; those open-circle points are not included in the graph. The value \(f(1)\) is not needed for this problem. a) Find \(\lim_{x\to 1^-}f(x)\). b) Find \(\lim_{x\to 1^+}f(x)\). c) Does \(\lim_{x\to 1}f(x)\) exist? Explain.
Figure for problem 551449

Hints

- An open circle marks a point that is not included in the graph, but its height can show what nearby values approach. - Trace the graph toward \(x=1\) from inputs smaller than \(1\), then from inputs larger than \(1\). - A finite two-sided limit exists only when the left-hand and right-hand limits agree.

Solution

a) Approaching \(x=1\) from the left, the graph stays at height \(2\), so \(\lim_{x\to 1^-}f(x)=2\). b) Approaching \(x=1\) from the right, the graph stays at height \(4\), so \(\lim_{x\to 1^+}f(x)=4\). c) The two one-sided limits are different, so the two-sided limit does not exist. The value of \(f(1)\), if one is assigned, would not change these one-sided limits.

Answer

a) \(2\) b) \(4\) c) The two-sided limit does not exist because the one-sided limits are unequal.
55145112
The graph shows \(f(x)=\sqrt{x}\), whose domain begins at \(x=0\). a) Find \(\lim_{x\to 0^+}f(x)\). b) Explain why a left-hand limit at \(0\) is not used for this real-valued function.
Figure for problem 551451

Hints

- Approach the endpoint using allowed input values. - Look at what happens to the graph as positive \(x\) gets closer to \(0\). - Check the domain before discussing a one-sided limit.

Solution

a) As positive \(x\)-values approach \(0\), the graph approaches \((0,0)\). Therefore, \(\lim_{x\to 0^+}f(x)=0\). b) The real-valued function has no domain values with \(x<0\), so there are no function values from the left that can approach \(0\).

Answer

a) \(0\) b) There are no real-domain inputs to the left of \(0\), so only the right-hand limit describes endpoint behavior.
53498712
The graph of \(g\) has a discontinuity at \(x=2\), marked by the vertical dashed asymptote. Determine from the graph whether \(\lim_{x\to2}g(x)\) exists. Justify your answer by describing the function values near the discontinuity.
Figure for problem 534987

Hints

- Examine the graph separately from the left and from the right of \(x=2\). - One-sided limits may be infinite. - A two-sided limit exists only when the two one-sided limits agree.

Solution

1. As \(x\to2^-\), the function values decrease without bound, so the left-hand limit is \(-\infty\). 2. As \(x\to2^+\), the function values increase without bound, so the right-hand limit is \(+\infty\). 3. A two-sided limit requires the one-sided limits to agree. Since they approach infinities with opposite signs, \(\lim_{x\to2}g(x)\) does not exist.

Answer

The limit does not exist because \(g(x)\to-\infty\) as \(x\to2^-\), while \(g(x)\to+\infty\) as \(x\to2^+\).
55145012
The graph represents one function \(f\). At \(x=1\), the open circle at \((1,1)\) is not included in the graph. The filled dot at \((1,3)\) is included and gives the actual function value, so \(f(1)=3\). a) Find \(\lim_{x\to 1^-}f(x)\) and \(\lim_{x\to 1^+}f(x)\). b) Find \(\lim_{x\to 1}f(x)\) and \(f(1)\). c) Is \(f\) continuous at \(x=1\)? Explain.
Figure for problem 551450

Hints

- An open circle marks a point that is not included in the graph, while a filled dot marks an included function value. - Read the height approached by the curve from each side of \(x=1\). - Continuity at a point requires the limit to exist and to equal the actual function value there.

Solution

a) From both sides, the curve approaches the open point at \((1,1)\), so \(\lim_{x\to 1^-}f(x)=1\) and \(\lim_{x\to 1^+}f(x)=1\). b) Since the one-sided limits agree, \(\lim_{x\to 1}f(x)=1\). The filled dot at \((1,3)\) is included, so \(f(1)=3\). c) The function is not continuous at \(x=1\) because the limit is \(1\) but the function value is \(3\).

Answer

a) \(\lim_{x\to 1^-}f(x)=1\) and \(\lim_{x\to 1^+}f(x)=1\) b) \(\lim_{x\to 1}f(x)=1\) and \(f(1)=3\) c) No. The limit exists, but it is not equal to \(f(1)\).
55145212
Let \(f(x)=x+2\) for \(x<1\), and \(f(x)=4-x\) for \(x\ge 1\). Find the left-hand limit, the right-hand limit, and the two-sided limit as \(x\to 1\). Then determine whether \(f\) is continuous at \(x=1\).

Hints

- Use the rule assigned to inputs less than \(1\) for the left-hand limit. - Use the rule assigned to inputs greater than \(1\) for the right-hand limit. - After comparing the one-sided limits, check the rule that actually includes \(x=1\).

Solution

1. For \(x<1\), use \(x+2\). As \(x\to 1^-\), this approaches \(3\). 2. For \(x\ge 1\), use \(4-x\). As \(x\to 1^+\), this approaches \(3\). 3. Because the one-sided limits agree, \(\lim_{x\to 1}f(x)=3\). 4. The second rule gives \(f(1)=3\), so the function is continuous at \(x=1\).

Answer

\(\lim_{x\to 1^-}f(x)=3\), \(\lim_{x\to 1^+}f(x)=3\), and \(\lim_{x\to 1}f(x)=3\). Also, \(f(1)=3\), so \(f\) is continuous at \(x=1\).
55145312
The graph shows \(f(x)=\frac{1}{(x-2)^2}\). a) Describe \(\lim_{x\to 2^-}f(x)\). b) Describe \(\lim_{x\to 2^+}f(x)\). c) What do these one-sided behaviors imply about \(\lim_{x\to 2}f(x)\) in the extended sense?
Figure for problem 551453

Hints

- Near \(x=2\), determine the sign and size of the squared denominator. - Compare what happens on the two sides of the vertical line \(x=2\). - Decide whether both branches grow in the same or opposite directions.

Solution

a) As \(x\) approaches \(2\) from the left, \((x-2)^2\) becomes a very small positive number, so \(f(x)\) grows without bound. Thus \(\lim_{x\to 2^-}f(x)=+\infty\). b) The same reasoning applies from the right, so \(\lim_{x\to 2^+}f(x)=+\infty\). c) Both sides grow without bound in the positive direction, so \(\lim_{x\to 2}f(x)=+\infty\) in the extended sense.

Answer

a) \(+\infty\) b) \(+\infty\) c) \(\lim_{x\to 2}f(x)=+\infty\) in the extended sense.
55145412
The table shows values of \(f(x)\) on both sides of \(x=2\). <table><tr><th>\(x<2\)</th><th>\(f(x)\)</th><th>\(x>2\)</th><th>\(f(x)\)</th></tr><tr><td>\(1.9\)</td><td>\(0.9\)</td><td>\(2.1\)</td><td>\(5.1\)</td></tr><tr><td>\(1.99\)</td><td>\(0.99\)</td><td>\(2.01\)</td><td>\(5.01\)</td></tr><tr><td>\(1.999\)</td><td>\(0.999\)</td><td>\(2.001\)</td><td>\(5.001\)</td></tr></table> Estimate the left-hand and right-hand limits at \(x=2\), and decide whether the two-sided limit exists.

Hints

- Read the left and right halves of the table separately. - On each side, focus on values from inputs closest to \(2\). - Compare the two target values before deciding whether a two-sided limit exists.

Solution

1. On the left, the outputs \(0.9\), \(0.99\), and \(0.999\) approach \(1\), so \(\lim_{x\to 2^-}f(x)=1\). 2. On the right, the outputs \(5.1\), \(5.01\), and \(5.001\) approach \(5\), so \(\lim_{x\to 2^+}f(x)=5\). 3. Since the one-sided limits are unequal, the two-sided limit does not exist.

Answer

\(\lim_{x\to 2^-}f(x)=1\), \(\lim_{x\to 2^+}f(x)=5\), and \(\lim_{x\to 2}f(x)\) does not exist.
55145512
A student examines a function defined by \(f(x)=x+1\) for \(x<0\), \(f(0)=10\), and \(f(x)=2-x\) for \(x>0\). The student makes two claims: Claim 1: “\(\lim_{x\to 0}f(x)=10\) because \(f(0)=10\).” Claim 2: “\(\lim_{x\to 0}f(x)=1\) because the left-hand limit is \(1\).” Explain why both claims are incorrect and state the correct conclusion.

Hints

- Treat the function value, left-hand behavior, and right-hand behavior as three separate pieces of information. - Find what each side approaches before considering a two-sided limit. - Ask what must be true about the two one-sided limits for a two-sided limit to exist.

Solution

1. Claim 1 is incorrect because a limit is determined by nearby values, not by the value at the single input \(x=0\). 2. From the left, \(x+1\) approaches \(1\), so \(\lim_{x\to 0^-}f(x)=1\). 3. From the right, \(2-x\) approaches \(2\), so \(\lim_{x\to 0^+}f(x)=2\). 4. Claim 2 is incorrect because a two-sided limit requires both one-sided limits to agree. Since \(1\ne 2\), \(\lim_{x\to 0}f(x)\) does not exist.

Answer

Both claims are incorrect. The function value \(f(0)=10\) does not determine the limit, and the left-hand limit alone is insufficient. Because the one-sided limits are \(1\) and \(2\), the two-sided limit does not exist.
55147512
The graph represents one function \(f\). All blue pieces and blue markers belong to that same function. An open circle marks a point that is not included in the graph; a filled dot marks the actual function value at that x-value. In particular, the graph shows \(f(-2)=2\) and \(f(1)=5\). The function has discontinuities at \(x=-2\) and \(x=1\). a) At which x-value could changing only the function value at that x-value make the function continuous? b) At which x-value would changing only the function value not be enough to make the function continuous? Explain each decision using the one-sided limits shown by the graph.
Figure for problem 551475

Hints

- Every blue piece and marker in the graph belongs to the same function \(f\). - At each marked x-value, first compare the heights approached from the left and from the right. - A filled dot gives the actual function value, but changing that one value does not change the nearby behavior on either side. - Changing one function value can repair continuity only when the two one-sided limits already agree.

Solution

a) At \(x=-2\), the graph approaches \(0\) from both sides, so \(\lim_{x\to -2^-}f(x)=\lim_{x\to -2^+}f(x)=0\). The filled dot shows that \(f(-2)=2\). Redefining only the function value to \(f(-2)=0\) would make the function continuous there. b) At \(x=1\), the graph approaches \(3\) from the left and \(5\) from the right. Because \(\lim_{x\to 1^-}f(x)=3\) and \(\lim_{x\to 1^+}f(x)=5\), the two-sided limit does not exist. Changing only \(f(1)\) cannot repair that jump.

Answer

a) \(x=-2\); redefining \(f(-2)=0\) would make the function continuous. b) \(x=1\); changing only \(f(1)\) cannot make the function continuous because the one-sided limits are \(3\) and \(5\).
55145612
Construct one piecewise-defined function \(f\) that satisfies all three conditions: \(\lim_{x\to 2^-}f(x)=-1\), \(\lim_{x\to 2^+}f(x)=3\), and \(f(2)=5\). Then state whether \(\lim_{x\to 2}f(x)\) exists and justify your answer.

Hints

- Treat inputs less than \(2\), equal to \(2\), and greater than \(2\) separately. - Choose behavior on each side that approaches the required target. - The assigned value at \(x=2\) can be chosen independently of nearby behavior. - Compare the two one-sided limits when deciding about the two-sided limit.

Solution

1. One valid construction is \(f(x)=-1\) for \(x<2\), \(f(2)=5\), and \(f(x)=3\) for \(x>2\). 2. Values to the left of \(2\) approach \(-1\), so the required left-hand limit is satisfied. 3. Values to the right of \(2\) approach \(3\), so the required right-hand limit is satisfied. The assigned value at \(x=2\) is \(5\). 4. Since the one-sided limits are unequal, the two-sided limit does not exist.

Answer

One possible function is \(f(x)=-1\) for \(x<2\), \(f(2)=5\), and \(f(x)=3\) for \(x>2\). The two-sided limit does not exist because \(-1\ne 3\).

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