Unit 1: Limits and Continuity

Students will explore how limits allow them to solve problems involving change and to better understand mathematical reasoning about functions.

Limits and Their Representations

Definition and Properties of Limits in Various Representations

The limit of a function \( \lim_{x \to a} f(x) = L \) means that as \( x \) gets arbitrarily close to \( a \), the values of \( f(x) \) get arbitrarily close to \( L \). This definition focuses on the behavior of \( f(x) \) near \( a \), not necessarily at \( a \) itself. For example, \( \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4 \) even though the function is undefined at \( x = 2 \).
Numerical representation uses a table of \( x \)-values approaching \( a \) from both sides, showing the corresponding \( f(x) \) values approaching \( L \). This is often used to estimate limits when algebraic manipulation is difficult, such as estimating \( \lim_{x\to 0} \frac{\sin x}{x} \) using decimal increments like 0.1, 0.01, 0.001.
Graphical representation involves observing the \( y \)-value that the function approaches as \( x \) nears \( a \) from both the left and right. If both sides agree, that is the limit. For example, on a graph of a piecewise function, you might see a hole at \( x = 3 \) but both sides approach \( y = 5 \), so the limit is 5.
Analytical representation uses algebra and known properties (limit laws) to calculate limits exactly. Key laws include: \( \lim_{x\to a} [f(x) \pm g(x)] = \lim_{x\to a} f(x) \pm \lim_{x\to a} g(x) \) and \( \lim_{x\to a} [f(x) \cdot g(x)] = (\lim_{x\to a} f(x)) \cdot (\lim_{x\to a} g(x)) \). These laws require that both individual limits exist.
Common mistakes include assuming a limit exists just because the function is defined at \( a \), or assuming that a vertical asymptote means the limit is zero. Always check left- and right-hand behavior, and remember that limits describe trends near a point, not necessarily the actual value at that point.

Limits: Numerical, Graphical, and Algebraic Evaluation Methods

Direct substitution works when \( f(x) \) is continuous at \( a \), meaning you can plug \( x = a \) into the formula to find \( L \). For example, \( \lim_{x\to 5} (3x + 2) = 17 \) because this linear function is continuous everywhere.
If direct substitution results in \( \frac{0}{0} \), you must use algebraic manipulation. For example, \( \lim_{x\to 2} \frac{x^2 - 4}{x - 2} \) simplifies by factoring to \( \lim_{x\to 2} (x + 2) = 4 \). This works because the factor \( (x - 2) \) causing the hole is removed.
Rationalizing is used for expressions with square roots that cause indeterminate forms. For example, \( \lim_{x\to 0} \frac{\sqrt{x + 4} - 2}{x} \) is simplified by multiplying numerator and denominator by \( \sqrt{x + 4} + 2 \) to eliminate the root from the numerator.
Graphical estimation involves tracing along the curve toward \( x = a \) and seeing what \( y \)-value is approached. This is helpful when the function is too complex to simplify algebraically but a graph is available.
Common pitfalls include forgetting to check both left and right limits separately when the function is piecewise, and misreading graphs when a hole and nearby values cause confusion. Always confirm that both sides approach the same number before concluding a limit exists.

One-Sided Limits and Piecewise Functions

One-Sided Limits

A left-hand limit \( \lim_{x \to a^-} f(x) \) considers only values of \( x \) slightly less than \( a \), while a right-hand limit \( \lim_{x \to a^+} f(x) \) considers only values slightly greater than \( a \). A limit \( \lim_{x \to a} f(x) \) exists only if both one-sided limits exist and are equal.
One-sided limits are essential for step functions or functions with jumps. For example, in the greatest integer function \( f(x) = \lfloor x \rfloor \), \( \lim_{x\to 2^-} f(x) = 1 \) but \( \lim_{x\to 2^+} f(x) = 2 \), so the overall limit at \( x = 2 \) does not exist.
When evaluating one-sided limits algebraically, you substitute values from only one side into the function. For rational functions, check the sign of the denominator as you approach the point to determine if the function tends toward \( +\infty \) or \( -\infty \).
Graphical analysis of one-sided limits focuses on the approach from one side of the graph. This is critical for identifying jumps, vertical asymptotes, or sudden changes in value at specific points.
Common errors include assuming symmetry in one-sided limits or ignoring the direction of approach. Always explicitly check each side to avoid incorrect conclusions about whether a limit exists.

Piecewise Functions and Limits

For a piecewise function, evaluating a limit at the boundary between two pieces requires checking the limit from both sides. This means using the rule for \( x < a \) to find \( \lim_{x \to a^-} f(x) \) and the rule for \( x > a \) to find \( \lim_{x \to a^+} f(x) \).
If both one-sided limits match, that is the limit of the piecewise function at the boundary. If they differ, the limit does not exist, even if the function is defined at that point.
Piecewise functions are often designed to be continuous, so parameters can be chosen to make limits match. For example, for \( f(x) = \begin{cases} kx + 1, & x < 2 \\ x^2 - k, & x \ge 2 \end{cases} \), continuity at \( x = 2 \) requires \( 2k + 1 = 4 - k \), giving \( k = 1 \).
When evaluating limits for piecewise functions graphically, follow each piece up to the boundary, then compare the \( y \)-values approached from each side. This avoids confusion from open circles and filled dots at boundaries.
Common mistakes include substituting into the wrong piece or skipping the separate one-sided evaluations, leading to incorrect assumptions about continuity or limit existence.

Squeeze Theorem and Special Trigonometric Limits

Squeeze Theorem

The Squeeze Theorem states that if \( g(x) \le f(x) \le h(x) \) for all \( x \) near \( a \) (except possibly at \( a \)), and \( \lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L \), then \( \lim_{x\to a} f(x) = L \). It’s used when \( f(x) \) is hard to evaluate directly but can be “trapped” between two simpler functions.
A classic example is \( \lim_{x\to 0} \frac{\sin x}{x} \). Knowing \( -1 \le \frac{\sin x}{x} \le 1 \) isn’t enough, but using geometry you can show \( \cos x \le \frac{\sin x}{x} \le 1 \) for \( x \) near 0, and since both bounds approach 1, the limit is 1.
When applying the theorem, clearly state the two bounding functions and verify the inequalities hold near \( a \). Skipping this step can lead to using invalid bounds and wrong conclusions.
The Squeeze Theorem also applies to oscillating functions like \( x \sin(1/x) \) as \( x \to 0 \). Since \( -|x| \le x \sin(1/x) \le |x| \) and both bounds approach 0, the limit is 0.
Common mistakes include forgetting to take the limit of both bounding functions or failing to prove the inequalities, which means the theorem’s conditions are not actually met.

Special Trigonometric Limits

The most important trig limit is \( \lim_{x\to 0} \frac{\sin x}{x} = 1 \), which is proven using the Squeeze Theorem and a unit circle geometry argument. This limit is foundational in calculus, especially in derivatives of trigonometric functions.
Another key limit is \( \lim_{x\to 0} \frac{1 - \cos x}{x} = 0 \), which can be found by multiplying numerator and denominator by \( 1 + \cos x \) and using the first trig limit to simplify.
These limits extend to forms involving \( ax \) instead of \( x \): \( \lim_{x\to 0} \frac{\sin(ax)}{x} = a \) and \( \lim_{x\to 0} \frac{1 - \cos(ax)}{x} = 0 \), derived by substitution \( u = ax \).
Memorizing these special trig limits is essential for differentiating trig functions and solving many AP Calculus BC limit problems efficiently. They are also commonly embedded in more complex algebraic expressions.
Common errors include confusing \( \frac{\sin x}{x} \) with \( \frac{x}{\sin x} \) (which approaches 1, not 0), and forgetting to adjust for coefficients inside the trig function during substitution.

Infinite Limits and Vertical Asymptotes

Infinite Limits

An infinite limit occurs when \( f(x) \) increases or decreases without bound as \( x \) approaches a certain value. For example, \( \lim_{x \to 0^+} \frac{1}{x} = +\infty \) because as \( x \) gets closer to zero from the right, the reciprocal gets larger and larger. In these cases, the limit is said to be infinite, not that it “doesn’t exist” — instead, we classify it as diverging to infinity.
Infinite limits often happen when the denominator of a rational function approaches zero while the numerator approaches a nonzero number. For example, \( \lim_{x \to 3} \frac{2}{x - 3} \) has different signs depending on whether we approach from the left or right, making one side approach \( -\infty \) and the other \( +\infty \).
To determine the sign of an infinite limit, test points just to the left and right of the value in question. This avoids guessing and ensures correct classification of the limit’s direction.
In AP problems, infinite limits appear in connection with asymptotic behavior and domain restrictions. Recognizing them is important for understanding function graphs, particularly in rational, trigonometric, and logarithmic functions.
Common mistakes include writing \( \infty \) as an actual limit value instead of saying “the limit diverges to infinity,” or failing to check both one-sided behaviors when the function is not symmetric.

Vertical Asymptotes

A vertical asymptote is a vertical line \( x = a \) where the function’s values increase or decrease without bound as \( x \) approaches \( a \). These occur when a factor in the denominator is zero but does not cancel with the numerator (non-removable discontinuity).
For rational functions, finding vertical asymptotes involves factoring numerator and denominator, canceling any common factors, and setting the remaining denominator factors equal to zero. For example, \( f(x) = \frac{x + 1}{(x - 2)(x + 3)} \) has vertical asymptotes at \( x = 2 \) and \( x = -3 \).
Some non-rational functions also have vertical asymptotes. For instance, \( \ln(x) \) has a vertical asymptote at \( x = 0 \) because it approaches \( -\infty \) as \( x \) approaches zero from the right.
When graphing, approach each vertical asymptote from both sides to determine whether the function goes to \( +\infty \) or \( -\infty \). This helps sketch an accurate graph and understand the function’s range.
Common mistakes include confusing holes (removable discontinuities) with vertical asymptotes. A hole occurs when a factor cancels, while a vertical asymptote remains after simplification.

Limits at Infinity and End Behavior

Limits at Infinity

Limits at infinity describe the value \( f(x) \) approaches as \( x \) increases or decreases without bound. This is critical for analyzing a function’s long-term behavior. For example, \( \lim_{x\to\infty} \frac{3x^2 + 5}{2x^2 - 7} = \frac{3}{2} \) because the highest-degree terms dominate.
For rational functions, compare the degree of the numerator and denominator: if numerator degree < denominator degree, the limit is 0; if degrees are equal, the limit is the ratio of leading coefficients; if numerator degree > denominator degree, the limit is \( \pm\infty \) depending on sign.
When \( x \to -\infty \), sign analysis is important. For example, \( \lim_{x \to -\infty} \frac{x^3}{x^2 + 1} \) diverges to \( -\infty \) because the leading term in the numerator dominates and is negative for large negative \( x \).
Other functions, such as exponential and logarithmic, have distinctive behaviors at infinity: \( e^x \) grows without bound as \( x\to\infty \), while \( \ln(x) \) grows slowly but unbounded as \( x\to\infty \).
Common mistakes include ignoring the sign of the highest-degree term when \( x \to -\infty \) and assuming the behavior for \( +\infty \) applies symmetrically to \( -\infty \).

End Behavior and Horizontal/Oblique Asymptotes

Horizontal asymptotes are horizontal lines \( y = L \) where \( \lim_{x\to\pm\infty} f(x) = L \). These show the output value the function settles toward for large magnitudes of \( x \).
If the numerator’s degree is exactly one more than the denominator’s, the function has an oblique (slant) asymptote found by polynomial long division. The quotient (without the remainder) gives the slant asymptote equation.
When determining end behavior, always reduce the function to simplest form and focus on the dominant term(s) for large \( |x| \). For example, \( \frac{5x^3 - 2x}{x^2 + 7} \) behaves like \( \frac{5x^3}{x^2} = 5x \), which increases without bound.
Connecting to earlier concepts: end behavior from limits at infinity complements vertical asymptote analysis from infinite limits, giving a full asymptotic picture of the graph.
Common mistakes include confusing horizontal asymptotes with restrictions on range — a function can cross its horizontal asymptote in the middle but will still approach it at infinity.

Continuity

Definition of Continuity of a Function at a Point and Over a Domain

A function \( f(x) \) is continuous at \( x = a \) if three conditions are met: \( f(a) \) is defined, \( \lim_{x\to a} f(x) \) exists, and \( \lim_{x\to a} f(x) = f(a) \). All three must hold — missing any means discontinuity at that point.
Continuity over a domain means the function is continuous at every point in that domain. For example, polynomials are continuous for all real numbers, while rational functions are continuous everywhere except where the denominator is zero.
Piecewise functions can be continuous if parameters are chosen to match limits at boundaries. This often appears in AP problems where you solve for a constant to make the function continuous.
Recognizing continuity types is important: removable discontinuity (hole), jump discontinuity, infinite discontinuity, and oscillating discontinuity all break the conditions of continuity differently.
Common mistakes include thinking a function is continuous at a point just because it is defined there, ignoring the necessity for the limit to equal the value, or forgetting to check both one-sided limits for piecewise boundaries.

Continuity and Discontinuities

Types of Discontinuities

Removable discontinuity occurs when the limit exists but the function’s value is either undefined or different from the limit. Example: \( f(x) = \frac{x^2 - 1}{x - 1} \) has a hole at \( x = 1 \) because the simplified form \( f(x) = x + 1 \) is defined everywhere except at \( x = 1 \). This type can be “fixed” by redefining the function at the missing point.
Jump discontinuity occurs when the left-hand and right-hand limits exist but are not equal. For example, a step function where \( f(x) = 2 \) for \( x < 0 \) and \( f(x) = 5 \) for \( x \ge 0 \) has a jump of 3 units at \( x = 0 \). These cannot be removed by changing a single point.
Infinite discontinuity occurs when the function approaches \( \pm\infty \) at a certain point. A classic example is \( f(x) = \frac{1}{x^2} \) at \( x = 0 \). This ties directly to vertical asymptotes and infinite limits discussed earlier.
Oscillating discontinuity occurs when a function oscillates infinitely as it approaches a point, making the limit nonexistent. Example: \( f(x) = \sin\left(\frac{1}{x}\right) \) as \( x \to 0 \) swings between -1 and 1 infinitely fast, so no limit exists.
Common mistakes include misidentifying removable discontinuities as vertical asymptotes or assuming a discontinuity is removable just because the function has a hole-like graph — always check if the limit exists and matches the value.

Intermediate Value Theorem

Understanding and Applying IVT

The Intermediate Value Theorem (IVT) states: If \( f \) is continuous on a closed interval \([a, b]\) and \( N \) is any number between \( f(a) \) and \( f(b) \), then there exists at least one \( c \in (a, b) \) such that \( f(c) = N \). This theorem guarantees solutions exist but does not find them.
To apply IVT, first confirm the function is continuous on the interval. Next, ensure the target value \( N \) lies between \( f(a) \) and \( f(b) \). Then conclude that a \( c \) exists in \((a, b)\) where \( f(c) = N \).
Example: For \( f(x) = x^3 - x - 2 \), check \( f(1) = -2 \) and \( f(2) = 4 \). Since \( 0 \) lies between -2 and 4, IVT tells us there is some \( c \in (1, 2) \) such that \( f(c) = 0 \) (a root of the equation).
IVT is often used to prove that equations have real solutions in a given interval, especially when exact roots are difficult to find algebraically. This connects to numerical approximation methods like bisection.
Common mistakes include using IVT on intervals where the function is not continuous (like a rational function across a vertical asymptote) or assuming it provides the exact value of \( c \) — it only proves existence.

Connections Between Continuity and Differentiability

How Continuity Relates to Differentiability

If a function is differentiable at \( x = a \), then it must also be continuous at \( x = a \). This is because the derivative’s limit definition requires the function’s value to match the limit from both sides. Differentiability implies continuity, but the reverse is not always true.
Continuity does not guarantee differentiability. For example, \( f(x) = |x| \) is continuous everywhere but not differentiable at \( x = 0 \) because the slopes from the left and right differ.
Points where a function is continuous but not differentiable include sharp corners (e.g., \( |x| \)), cusps (e.g., \( x^{2/3} \) at \( x = 0 \)), and vertical tangents (e.g., \( \sqrt[3]{x} \) at \( x = 0 \) has infinite slope).
This connection is important when analyzing motion: a continuous position function can still have “sudden changes in velocity” if it’s not differentiable at certain times, which would correspond to instantaneous directional changes.
Common mistakes include thinking differentiability can fix discontinuity or forgetting that differentiability is a stronger condition — if a function has any discontinuity, it cannot be differentiable there.

Practice Problems

Problem 1 (Continuity of a Piecewise Function)

Question. Consider \[ f(x)= \begin{cases} \dfrac{x^2-4}{x-2}, & x<2,\\[4pt] kx+m, & x\ge 2. \end{cases} \] (a) Compute \( \lim_{x\to 2^-} f(x) \). (b) Find constants \( k \) and \( m \) so that \( f \) is continuous at \( x=2 \) and satisfies \( f(2)=5 \). (c) State the type of discontinuity at \( x=2 \) if continuity is not enforced and explain why.
Step 1: Simplify the left piece to evaluate the left-hand limit. For \( x<2 \), factor the numerator \( x^2-4=(x-2)(x+2) \) so \( \dfrac{x^2-4}{x-2}=x+2 \) for \( x\neq 2 \). Therefore \( \lim_{x\to 2^-} f(x)=\lim_{x\to 2^-} (x+2)=4 \), which comes from direct substitution into the simplified expression because limits ignore the single excluded point. This identifies the target value the right-hand expression must match to secure continuity at \( x=2 \).
Step 2: Impose the continuity condition using the right-hand rule. For continuity at \( x=2 \), we need \( \lim_{x\to 2^-} f(x)=\lim_{x\to 2^+} f(x)=f(2) \). The right-hand expression is \( kx+m \), so \( \lim_{x\to 2^+} f(x)=2k+m \). Matching the left-hand limit gives \( 2k+m=4 \), which is the continuity equation linking \( k \) and \( m \).
Step 3: Use the value constraint \( f(2)=5 \) to solve for \( k \) and \( m \). Because the second branch defines \( f \) at \( x=2 \), we have \( f(2)=2k+m=5 \). Together with the continuity equation \( 2k+m=4 \), we get a linear system: \[ \begin{cases} 2k+m=4,\\ 2k+m=5. \end{cases} \] This system is inconsistent, which means you cannot simultaneously satisfy both “continuous at \( x=2 \)” and “\( f(2)=5 \)” with a linear \( kx+m \). Therefore, no such \( k,m \) exist.
Step 4: Interpret the result and answer each part clearly. (a) \( \lim_{x\to 2^-} f(x)=4 \). (b) There are no \( k,m \) that make \( f \) continuous at \( x=2 \) while also enforcing \( f(2)=5 \); the best you can do for continuity is require \( f(2)=4 \), not \( 5 \). (c) If you set \( f(2)=5 \) anyway, you create a removable discontinuity (a hole) at \( y=4 \) with a mismatched filled point at \( (2,5) \), because the two-sided limit exists and equals \( 4 \) but the function’s value is \( 5 \).
Common mistakes and connections. A frequent error is plugging \( x=2 \) into \( \dfrac{x^2-4}{x-2} \) before simplifying and claiming “undefined means no limit,” which is false because the limit uses nearby values and equals \( 4 \). Another mistake is assuming any target value for \( f(2) \) can be achieved while keeping continuity with a single linear piece; the limit forces \( f(2) \) to match \( 4 \). This problem connects definitions (limit and continuity), one-sided limits, and the distinction between function values and limiting behavior.

Problem 2 (Limits at Infinity, Asymptotes, and One-Sided Behavior)

Question. Let \( g(x)=\dfrac{3x^2-2x+1}{x^2-4x-5} \). (a) Compute \( \lim_{x\to\infty} g(x) \) and \( \lim_{x\to -\infty} g(x) \). (b) Identify any horizontal or oblique asymptotes. (c) Find the vertical asymptotes and determine the one-sided behavior near each. Present clear, sign-based reasoning in each step.
Step 1: Determine end behavior via degree comparison. The numerator and denominator are both degree 2, so the limits at \( \pm\infty \) equal the ratio of leading coefficients. Therefore \( \lim_{x\to\pm\infty} g(x)=\dfrac{3}{1}=3 \). This immediately yields a horizontal asymptote \( y=3 \) and rules out an oblique asymptote because the degrees are equal, not numerator one higher.
Step 2: Factor the denominator to locate vertical asymptotes. Solve \( x^2-4x-5=0 \Rightarrow (x-5)(x+1)=0 \), giving candidates \( x=5 \) and \( x=-1 \). Since no common factors cancel with the numerator (check quickly that \( 3x^2-2x+1 \) shares neither \( x-5 \) nor \( x+1 \)), both produce vertical asymptotes. This confirms non-removable discontinuities at these \( x \)-values.
Step 3: Analyze one-sided behavior near \( x=5 \). Take a value just to the left, say \( x=5^- \), then \( x-5<0 \) and \( x+1>0 \), so the denominator \( (x-5)(x+1) \) is negative. At \( x\approx 5 \), the numerator \( 3x^2-2x+1 \) is positive (\( 3\cdot25-10+1=66>0 \)), so the fraction is positive over negative, which tends to \( -\infty \). For \( x=5^+ \), the denominator becomes positive and the numerator stays positive, so \( g(x)\to +\infty \).
Step 4: Analyze one-sided behavior near \( x=-1 \). For \( x=-1^- \), \( x-5<0 \) and \( x+1<0 \), so the denominator is product of two negatives, hence positive; the numerator at \( x\approx -1 \) is \( 3(1)+2+1=6>0 \), so \( g(x)\to +\infty \). For \( x=-1^+ \), the factor \( x+1>0 \) while \( x-5<0 \), making the denominator negative and keeping the numerator positive, so \( g(x)\to -\infty \). This completes the sign chart and the asymptotic sketch near each vertical line.
Step 5: State the final results clearly and connect concepts. (a) \( \lim_{x\to\infty} g(x)=3 \) and \( \lim_{x\to -\infty} g(x)=3 \). (b) Horizontal asymptote: \( y=3 \); no oblique asymptote. (c) Vertical asymptotes: \( x=5 \) with \( g(x)\to -\infty \) from the left and \( +\infty \) from the right; \( x=-1 \) with \( g(x)\to +\infty \) from the left and \( -\infty \) from the right; these behaviors follow from factoring and sign analysis.
Common mistakes and quick checks. A common error is to cancel non-existent common factors or to forget that equal degrees give a horizontal asymptote by lead-coefficient ratio. Another frequent mistake is guessing the one-sided behavior without a systematic sign chart; always evaluate factors’ signs just to the left and right of each asymptote. This problem ties together limits at infinity, asymptotes, discontinuities, and one-sided limits into a coherent graph-based understanding.