Aimathic
Login | English | Deutsch

Free math worksheets

Build your own math worksheets from 30,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

One-sided limits and discontinuities

Click problems to add them to your worksheet.

55144912
The graph represents one function \(f\) near \(x=1\). 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

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.
55145112
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.
55574412
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
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
Use the graph of \(f\). 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 \(f(1)\).
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
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.
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.
55574512
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 \(x=2\).
55145512
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.
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 \(5\).
55574612
For \(x\ne0\), let \(f(x)=\sin\left(\frac{1}{x}\right)\). The graph shows its behavior near \(x=0\). a) Does \(\lim_{x\to0^+}f(x)\) exist? b) Does \(\lim_{x\to0}f(x)\) exist? If not, explain what prevents a limit.
Figure for problem 555746

Hints

- Focus on whether the graph settles toward one height as it gets closer to \(x=0\). - Think about what happens to \(1/x\) as positive \(x\) approaches \(0\). - A limit requires nearby function values to approach a single number.

Solution

1. As positive \(x\) values approach \(0\), \(\frac{1}{x}\) grows without bound and the sine values keep oscillating between \(-1\) and \(1\). 2. The function therefore does not approach one number from the right, so \(\lim_{x\to0^+}f(x)\) does not exist. 3. Since even the right-hand limit fails to exist, the two-sided limit \(\lim_{x\to0}f(x)\) also does not exist.

Answer

a) No. b) No. The function keeps oscillating rather than approaching one value as \(x\to0\).
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 \(-1\ne 3\).

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.

Building free math worksheets