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One-sided limits and discontinuities

Learning goals

  • Determines left- and right-hand limits.
  • Uses agreement of one-sided limits to decide whether a two-sided limit exists.
  • Classifies discontinuity behavior from limits and function values.
  • Distinguishes jump, removable, and infinite behavior informally.

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One-sided limits and discontinuities 1
Use the graph to compare the values approached from each side of \(x=1\). 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

- Trace the graph toward \(x=1\) separately from the left and the right. - Compare the heights being approached, not the value at the input itself. - What must agree for a two-sided limit to exist?

Solution

1. Approaching \(x=1\) from the left, the graph stays at height \(2\), so \(\lim_{x\to 1^-}f(x)=2\). 2. Approaching \(x=1\) from the right, the graph stays at height \(4\), so \(\lim_{x\to 1^+}f(x)=4\). 3. 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.
One-sided limits and discontinuities 2
Use the graph of the real-valued function \(f\). a) Find \(\lim_{x\to0^+}f(x)\). b) Explain why a left-hand limit at \(0\) is not used for the function shown.
Figure for problem 551451

Hints

- Approach the endpoint using only inputs where the graph exists. - Read the height approached as \(x\) moves toward \(0\) from the right. - Check whether the graph has any real-domain points to the left of the endpoint.

Solution

1. As allowed inputs approach \(0\) from the right, the graph approaches height \(0\). Therefore, \(\lim_{x\to0^+}f(x)=0\). 2. The graph has no real-domain points 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 the endpoint behavior.
One-sided limits and discontinuities 3
Match each local behavior with the standard discontinuity type: removable, jump, or infinite. a) As \(x\to2\), the function approaches \(4\) from both sides, but \(f(2)=1\). b) As \(x\to-1\), the left-hand limit is \(2\) and the right-hand limit is \(5\). c) As \(x\to3\), the function values increase without bound from both sides.

Hints

- Ask first whether the two one-sided limits agree. - If both sides approach the same finite value, consider whether changing only the value at the point would repair continuity. - If values grow without bound near the input, identify the corresponding discontinuity type.

Solution

1. In a), both one-sided limits agree at a finite value, and only the value at the point prevents continuity. This is removable. 2. In b), the two one-sided limits are finite but unequal. This is a jump discontinuity. 3. In c), the function grows without bound near the input. This is an infinite discontinuity.

Answer

a) removable b) jump c) infinite
One-sided limits and discontinuities 4
Write the notation for the limit of \(f(x)\) as \(x\) approaches \(2\) from the left. What does the small sign on \(2\) mean?

Hints

- Decide which side of \(2\) contains inputs less than \(2\). - The small sign records the side from which the input approaches, not the sign of the output.

Solution

1. Approaching from the left means using inputs smaller than \(2\). 2. Write \(\lim_{x\to2^-}f(x)\). The superscript minus sign shows the side of approach.

Answer

\(\lim_{x\to2^-}f(x)\). The minus sign means inputs less than \(2\).
One-sided limits and discontinuities 5
The table gives values of \(g(x)\) near \(x=0\). <table><tr><th>\(x\)</th><td>\(-0.1\)</td><td>\(-0.01\)</td><td>\(0.01\)</td><td>\(0.1\)</td></tr><tr><th>\(g(x)\)</th><td>\(2.9\)</td><td>\(2.99\)</td><td>\(4.01\)</td><td>\(4.1\)</td></tr></table> Estimate the left-hand and right-hand limits at \(0\).

Hints

- Separate the negative inputs from the positive inputs. - Find the nearby target number for each group independently.

Solution

1. From negative inputs, the outputs approach \(3\). 2. From positive inputs, the outputs approach \(4\).

Answer

Left-hand limit: approximately \(3\). Right-hand limit: approximately \(4\).
One-sided limits and discontinuities 6
Does the function have a two-sided limit as \(x\to1\)? Explain from the graph.
Figure for problem 561479

Hints

- Follow the branch on either side toward the same input. - A two-sided limit requires the two heights to agree.

Solution

1. From the left, the graph approaches height \(-1\). 2. From the right, it approaches height \(2\). 3. The two approaches disagree, so the two-sided limit does not exist.

Answer

No. The left-hand limit is \(-1\) and the right-hand limit is \(2\).
One-sided limits and discontinuities 7
A movie theater sells tickets for \(\$10\) before 5 p.m. and \(\$12\) starting at 5 p.m. A student says the ticket price has a two-sided limit as the purchase time approaches 5 p.m., because the price stays constant on each side. Is that right?

Hints

- Consider a purchase a few seconds before the price change and one a few seconds after. - Does a two-sided limit depend on each side being constant, or on the two sides agreeing?

Solution

1. Just before 5 p.m., the price approaches \(\$10\); just after, it approaches \(\$12\). 2. Since these one-sided limits differ, no two-sided limit exists despite each price being constant on its side.

Answer

No. The price approaches \(\$10\) from before 5 p.m. and \(\$12\) from after 5 p.m., so there is no two-sided limit.
One-sided limits and discontinuities 8
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 \(x\to2^+\).
One-sided limits and discontinuities 9
a) Find \(\lim_{x\to1^-}f(x)\) and \(\lim_{x\to1^+}f(x)\). b) Find \(\lim_{x\to1}f(x)\) and \(f(1)\). c) Is \(f\) continuous at \(x=1\)? Explain.
Figure for problem 551450

Hints

- Trace the curve toward \(x=1\) from the left and from the right. - Use the filled point only for the actual function value. - Continuity requires the two-sided limit and the function value to agree.

Solution

1. From both sides of \(x=1\), the curve approaches height \(1\). Thus \(\lim_{x\to1^-}f(x)=1\) and \(\lim_{x\to1^+}f(x)=1\). 2. Because the one-sided limits agree, \(\lim_{x\to1}f(x)=1\). The filled point at \(x=1\) has height \(3\), so \(f(1)=3\). 3. The function is not continuous at \(x=1\) because its limit is \(1\) but its function value is \(3\).

Answer

a) \(\lim_{x\to1^-}f(x)=1\) and \(\lim_{x\to1^+}f(x)=1\) b) \(\lim_{x\to1}f(x)=1\) and \(f(1)=3\) c) No. The limit exists, but it is not equal to \(f(1)\).
One-sided limits and discontinuities 10
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 \(x=1\).
One-sided limits and discontinuities 11
Use the graph of \(f\). a) Describe \(\lim_{x\to2^-}f(x)\). b) Describe \(\lim_{x\to2^+}f(x)\). c) What do these one-sided behaviors imply about \(\lim_{x\to2}f(x)\) in the extended sense?
Figure for problem 551453

Hints

- Trace each branch toward the vertical line \(x=2\). - Decide whether the function values increase or decrease without bound on each side. - Compare the two one-sided behaviors before stating the two-sided extended limit.

Solution

1. As \(x\) approaches \(2\) from the left, the graph rises without bound, so \(\lim_{x\to2^-}f(x)=+\infty\). 2. The graph also rises without bound as \(x\) approaches \(2\) from the right, so \(\lim_{x\to2^+}f(x)=+\infty\). 3. Because both one-sided limits are \(+\infty\), \(\lim_{x\to2}f(x)=+\infty\) in the extended sense.

Answer

a) \(+\infty\) b) \(+\infty\) c) \(\lim_{x\to2}f(x)=+\infty\) in the extended sense.
One-sided limits and discontinuities 12
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.
One-sided limits and discontinuities 13
Let \(f(x)=x+1\) for \(x<2\), and \(f(x)=x^2\) for \(x\ge2\). a) Find \(\lim_{x\to2^-}f(x)\). b) Find \(\lim_{x\to2^+}f(x)\). c) Does \(\lim_{x\to2}f(x)\) exist? Classify the discontinuity.

Hints

- For a left-hand limit, use the formula assigned to inputs less than \(2\). - For a right-hand limit, use the formula assigned to inputs at least \(2\). - Compare the two one-sided limits before deciding whether a two-sided limit exists.

Solution

1. From the left, use \(x+1\): \(\lim_{x\to2^-}f(x)=3\). 2. From the right, use \(x^2\): \(\lim_{x\to2^+}f(x)=4\). 3. Because the one-sided limits are finite but unequal, the two-sided limit does not exist and the function has a jump discontinuity at \(x=2\).

Answer

a) \(3\) b) \(4\) c) The two-sided limit does not exist; there is a jump discontinuity at \(x=2\).
One-sided limits and discontinuities 14
Is the discontinuity at \(x=0\) removable or a jump? Explain using the left- and right-hand limits.
Figure for problem 561480

Hints

- Compare where the straight branch ends with where the curved branch begins. - Would moving just the filled point make the two-sided approach and the function value agree?

Solution

1. The left straight branch and right curved branch both approach height \(2\) as \(x\to0\). 2. The filled point shows a different function value, \(f(0)=4\). 3. Changing that single function value would remove the discontinuity, so it is removable.

Answer

Removable discontinuity: both sides approach \(2\), but \(f(0)=4\).
One-sided limits and discontinuities 15
Let \(f(x)=\frac{1}{x-2}\). From which side of \(x=2\) do the function values grow without bound in the positive direction? What happens from the other side?
Figure for problem 561481

Hints

- Compare the signs of \(x-2\) on either side of \(2\). - What happens when \(1\) is divided by a number very close to zero?

Solution

1. For \(x>2\), the denominator is small and positive near \(2\), so the outputs grow toward positive infinity. 2. For \(x<2\), the denominator is small and negative, so the outputs decrease without bound.

Answer

From the right, the outputs grow toward \(+\infty\); from the left, they decrease toward \(-\infty\).
One-sided limits and discontinuities 16
The curve would be continuous if one plotted point were moved vertically. What should the function value \(f(1)\) be?
Figure for problem 561482

Hints

- Trace the line toward the missing input from both sides. - Moving a filled point changes the function value there, but does it change the nearby line?

Solution

1. As \(x\) approaches \(1\), both branches approach height \(2\). 2. The value at \(x=1\) must equal that approached height for continuity.

Answer

\(f(1)=2\)
One-sided limits and discontinuities 17
Giulia examines a function defined by \(f(x)=x+1\) for \(x<0\), \(f(0)=10\), and \(f(x)=2-x\) for \(x>0\). Giulia 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.
One-sided limits and discontinuities 18
The function has discontinuities at \(x=-2\) and \(x=1\). a) At which x-value could changing only the function value make the function continuous? b) At which x-value would changing only the function value not be enough? Explain using the one-sided limits.
Figure for problem 551475

Hints

- At each marked input, compare the heights approached from either side. - Changing the filled-point value cannot change the heights approached by nearby graph points. - A single-value repair works only when the one-sided limits already match.

Solution

1. At \(x=-2\), both sides approach \(0\), but the filled dot gives \(f(-2)=2\). Changing only \(f(-2)\) to \(0\) would restore continuity. 2. At \(x=1\), the left-hand limit is \(3\) and the right-hand limit is \(5\). The two-sided limit does not exist, so changing only \(f(1)\) cannot repair the jump.

Answer

a) \(x=-2\); setting \(f(-2)=0\) would make the function continuous. b) \(x=1\); the one-sided limits \(3\) and \(5\) are unequal.
One-sided limits and discontinuities 19
Let \(f(x)=\sin\left(\frac{1}{x}\right)\) for \(x\ne0\). a) Does \(\lim_{x\to0^+}f(x)\) exist? Explain. b) Does \(\lim_{x\to0}f(x)\) exist? Explain.

Hints

- What happens to \(\frac{1}{x}\) when positive \(x\) gets closer to zero? - Think about the values sine takes as its input keeps growing. - What must happen from the right for a two-sided limit to exist?

Solution

1. As positive \(x\) approaches \(0\), \(\frac{1}{x}\) increases without bound. 2. Sine continues to oscillate between \(-1\) and \(1\) as its input grows, so \(f(x)\) does not approach a single value from the right. 3. The right-hand limit does not exist. Therefore, the two-sided limit does not exist either.

Answer

a) No. As \(x\to0^+\), the sine values keep oscillating rather than approaching one number. b) No. A two-sided limit cannot exist when the right-hand limit does not exist.
One-sided limits and discontinuities 20
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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