Limits & Continuity | Free AP Calculus AB Course

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<title>AP Calculus AB: Limits and Continuity | The School of Mathematics</title>

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<h1>AP Calculus AB: Limits and Continuity</h1>

Limits are the foundation everything else in calculus is built on, derivatives, integrals, and continuity itself are all defined in terms of them. This free, complete lesson covers one-sided limits, infinite limits and end behavior, limit theorems and algebraic techniques, continuity and the types of discontinuities, and the Extreme Value and Intermediate Value Theorems. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the full AP Calculus AB Question Bank linked below.

</p>

<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/Limits%26Continuity-Quiz%201">Practice Limits & Continuity Free</a>

<a class="cta-btn cta-secondary" href="https://theschoolofmathematics.com/quiz/course/AP-Calculus-AB-QBank">Explore the Full AP Calculus AB Qbank</a>

</div>

<h2>What's covered in this lesson</h2>

<ol>

<li><a href="#one-sided">One-Sided Limits and the Existence of a Limit</a></li>

<li><a href="#infinite-limits">Infinite Limits and End Behavior</a></li>

<li><a href="#techniques">Limit Theorems and Algebraic Techniques</a></li>

<li><a href="#continuity">Continuity and Types of Discontinuities</a></li>

<li><a href="#evt-ivt">The Extreme Value and Intermediate Value Theorems</a></li>

<li><a href="#mistakes">Common Mistakes to Avoid</a></li>

<li><a href="#faq">Frequently Asked Questions</a></li>

</ol>

</nav>

<h2 id="one-sided">1. One-Sided Limits and the Existence of a Limit</h2>

<p>The number L is the <strong>limit</strong> of f(x) as x approaches c if, as x gets arbitrarily close to c (without equaling c), f(x) gets arbitrarily close to L. We write lim<sub>x→c</sub> f(x) = L. The <strong>left-hand limit</strong>, lim<sub>x→c⁻</sub> f(x), only considers x-values less than c; the <strong>right-hand limit</strong>, lim<sub>x→c⁺</sub> f(x), only considers x-values greater than c. The two-sided limit exists only when both one-sided limits exist and agree.</p>

<p><strong>Worked Example:</strong> For the greatest integer function g(x) = &lfloor;x&rfloor;, find lim<sub>x→5⁻</sub> g(x) and lim<sub>x→5⁺</sub> g(x). Does lim<sub>x→5</sub> g(x) exist?</p>

<p>Approaching 5 from the left (values like 4.9, 4.99), &lfloor;x&rfloor; = 4. Approaching from the right (values like 5.01, 5.1), &lfloor;x&rfloor; = 5.</p>

<p>Since the one-sided limits (4 and 5) disagree, lim<sub>x→5</sub> g(x) does not exist.</p>

</div>

<p class="prompt">1. For the greatest integer function g(x) = &lfloor;x&rfloor;, what is lim<sub>x→2⁻</sub> g(x)?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pa-1',false)">D) Does not exist</button>

</div>

<p><span class="label">Explanation:</span> Approaching 2 from the left (like 1.9, 1.99), &lfloor;x&rfloor; = 1.</p>

</div>

<p class="prompt">2. For the same function, what is lim<sub>x→2⁺</sub> g(x)?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pa-2',false)">D) Does not exist</button>

</div>

<p><span class="label">Explanation:</span> Approaching 2 from the right (like 2.01, 2.1), &lfloor;x&rfloor; = 2.</p>

</div>

<p class="prompt">3. Based on the previous two results, does lim<sub>x→2</sub> g(x) exist?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pa-3',true)">A) No, since the left- and right-hand limits differ</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pa-3',false)">B) Yes, it equals 1</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pa-3',false)">C) Yes, it equals 2</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pa-3',false)">D) Yes, it equals 1.5</button>

</div>

<p><span class="label">Explanation:</span> A two-sided limit exists only when both one-sided limits agree. Here they're 1 and 2, so the limit does not exist.</p>

</div>

<p class="prompt">4. A function f is defined by f(x) = x + 2 for x < 1, and f(x) = 5 − 2x for x ≥ 1. Does lim<sub>x→1</sub> f(x) exist?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pa-4',true)">A) Yes, it equals 3</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pa-4',false)">B) No, it does not exist</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pa-4',false)">C) Yes, it equals 1</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pa-4',false)">D) Yes, it equals 5</button>

</div>

<p><span class="label">Explanation:</span> Left-hand: lim<sub>x→1⁻</sub>(x + 2) = 3. Right-hand: lim<sub>x→1⁺</sub>(5 − 2x) = 3. Since both agree, the limit exists and equals 3, even though this is a piecewise boundary point.</p>

</div>

<h2 id="infinite-limits">2. Infinite Limits and End Behavior</h2>

<p>A function <strong>becomes infinite</strong> as x approaches c if f(x) can be made arbitrarily large (positive or negative) by taking x sufficiently close to c. Writing lim<sub>x→c</sub> f(x) = +∞ or −∞ describes this behavior, it does not mean the limit exists, since a limit must be a finite number. For end behavior, every polynomial of degree ≥ 1 becomes infinite as x does, with the sign determined by the leading coefficient and whether the degree is even or odd.</p>

<p><strong>Worked Example:</strong> Describe the behavior of f(x) = <span class="frac"><span class="num">1</span><span class="den">x − 3</span></span> near x = 3.</p>

<p>As x → 3⁻ (like x = 2.9), the denominator is a small negative number, so f(x) → −∞.</p>

<p>As x → 3⁺ (like x = 3.1), the denominator is a small positive number, so f(x) → +∞.</p>

<p class="step-math">Since the one-sided behaviors disagree, lim<sub>x→3</sub> f(x) does not exist.</p>

</div>

<p class="prompt">1. Describe the behavior of f(x) = <span class="frac"><span class="num">1</span><span class="den">x + 2</span></span> as x → −2⁻.</p>

</div>

<p><span class="label">Explanation:</span> As x → −2 from the left (like −2.1), the denominator is a small negative number, so f(x) → −∞.</p>

</div>

<p class="prompt">2. Describe the behavior of the same function as x → −2⁺.</p>

</div>

<p><span class="label">Explanation:</span> As x → −2 from the right (like −1.9), the denominator is a small positive number, so f(x) → +∞.</p>

</div>

</div>

<p><span class="label">Explanation:</span> The leading term −3x⁵ dominates. Since the degree is odd and the coefficient is negative, as x → +∞, the function → −∞.</p>

</div>

</div>

<p><span class="label">Explanation:</span> As x → −∞, x⁵ → −∞ (an odd power of a very negative number is very negative). Multiplying by −3 flips the sign, giving +∞.</p>

</div>

<p class="prompt">5. The graph of h(x) = <span class="frac"><span class="num">x + 3</span><span class="den">x − 5</span></span> has a horizontal asymptote at what line?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pb-5',false)">D) There is no horizontal asymptote</button>

</div>

<p><span class="label">Explanation:</span> The numerator and denominator have the same degree (both 1), so the horizontal asymptote is the ratio of their leading coefficients: <span class="frac"><span class="num">1</span><span class="den">1</span></span> = 1.</p>

</div>

<h2 id="techniques">3. Limit Theorems and Algebraic Techniques</h2>

<p>Limits distribute over sums, differences, products, and quotients (provided the denominator's limit isn't 0), which justifies substituting directly whenever a function is continuous at the point in question. When direct substitution gives <span class="frac"><span class="num">0</span><span class="den">0</span></span>, factor and cancel first. The <strong>Squeeze (Sandwich) Theorem</strong> says that if f(x) ≤ g(x) ≤ h(x) and f and h share the same limit L at some point, then g must share that limit too. The basic trigonometric limit lim<sub>θ→0</sub> <span class="frac"><span class="num">sin θ</span><span class="den">θ</span></span> = 1 (θ in radians) is a direct consequence of the Squeeze Theorem.</p>

<p><strong>Worked Example:</strong> Find lim<sub>x→4</sub> <span class="frac"><span class="num">x² − 16</span><span class="den">x − 4</span></span>.</p>

</div>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pc-1',false)">D) Does not exist</button>

</div>

<p><span class="label">Explanation:</span> <span class="frac"><span class="num">(x − 2)(x + 2)</span><span class="den">x − 2</span></span> = x + 2, so the limit as x → 2 is 4.</p>

</div>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pc-2',false)">D) Does not exist</button>

</div>

<p><span class="label">Explanation:</span> x² + 2x − 3 factors as (x + 3)(x − 1), so <span class="frac"><span class="num">(x + 3)(x − 1)</span><span class="den">x + 3</span></span> = x − 1. At x = −3: −3 − 1 = −4.</p>

</div>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pc-3',true)">A) Does not exist</button>

</div>

<p><span class="label">Explanation:</span> Simplify: <span class="frac"><span class="num">4x² − 3x</span><span class="den">x²</span></span> = 4 − <span class="frac"><span class="num">3</span><span class="den">x</span></span>. As x → 0, <span class="frac"><span class="num">3</span><span class="den">x</span></span> becomes infinite with opposite signs from each side, so the limit does not exist.</p>

</div>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pc-4',false)">B) Cannot be determined</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pc-4',false)">D) It depends on the specific function g</button>

</div>

<p><span class="label">Explanation:</span> By the Squeeze Theorem, since g is trapped between two functions that both approach 0, g must also approach 0.</p>

</div>

</div>

<p><span class="label">Explanation:</span> Rewrite as 5 · <span class="frac"><span class="num">sin(5x)</span><span class="den">5x</span></span>. Since 5x → 0 as x → 0, the fraction approaches 1, leaving 5 · 1 = 5.</p>

</div>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pc-6',false)">C) Does not exist</button>

</div>

<p><span class="label">Explanation:</span> Since −1 ≤ sin x ≤ 1 for all x, we have <span class="frac"><span class="num">−1</span><span class="den">x</span></span> ≤ <span class="frac"><span class="num">sin x</span><span class="den">x</span></span> ≤ <span class="frac"><span class="num">1</span><span class="den">x</span></span>. Both outer bounds approach 0 as x → ∞, so by the Squeeze Theorem, the limit is 0.</p>

</div>

<h2 id="continuity">4. Continuity and Types of Discontinuities</h2>

<p>A function y = f(x) is <strong>continuous</strong> at x = c if (1) f(c) exists, (2) lim<sub>x→c</sub> f(x) exists, and (3) lim<sub>x→c</sub> f(x) = f(c). Polynomials are continuous everywhere; rational functions are continuous everywhere except where the denominator is 0.</p>

<tr><th>Discontinuity type</th><th>What's happening</th><th>Removable?</th></tr>

<tr><td>Removable</td><td>The limit exists, but either f(c) doesn't exist or doesn't match the limit</td><td>Yes, by redefining f(c)</td></tr>

<tr><td>Jump</td><td>The left- and right-hand limits both exist but disagree</td><td>No</td></tr>

<tr><td>Infinite</td><td>The function has a vertical asymptote at x = c</td><td>No</td></tr>

</table>

<p><strong>Worked Example:</strong> Is f(x) = <span class="frac"><span class="num">x² − 9</span><span class="den">x − 3</span></span> (x ≠ 3) continuous at x = 3? If not, is the discontinuity removable?</p>

<p>f is not defined at x = 3, so it fails condition (1) and is not continuous there.</p>

<p>However, lim<sub>x→3</sub> <span class="frac"><span class="num">(x − 3)(x + 3)</span><span class="den">x − 3</span></span> = lim<sub>x→3</sub> (x + 3) = 6 does exist.</p>

<p>Since the limit exists but f(3) doesn't, this discontinuity is removable: redefining f(3) = 6 would make f continuous there.</p>

</div>

<p class="prompt">1. Is f(x) = 3x² − 2x + 5 continuous at x = −1?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-1',false)">C) Cannot be determined</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-1',false)">D) Only if evaluated numerically</button>

</div>

<p><span class="label">Explanation:</span> Polynomials are continuous at every real number, including x = −1.</p>

</div>

<p class="prompt">2. Is g(x) = <span class="frac"><span class="num">1</span><span class="den">x + 4</span></span> continuous at x = −4?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-2',false)">C) Cannot be determined</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-2',false)">D) Yes, but only removably</button>

</div>

<p><span class="label">Explanation:</span> The denominator is 0 at x = −4, so g is undefined there. This is an infinite discontinuity, not continuous.</p>

</div>

<p class="prompt">3. Is k(x) = <span class="frac"><span class="num">x² − 4</span><span class="den">x − 2</span></span> (x ≠ 2) continuous at x = 2, and if not, is the discontinuity removable?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pd-3',true)">A) Not continuous, but the discontinuity is removable</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-3',false)">B) Continuous at x = 2</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-3',false)">C) Not continuous, and the discontinuity is not removable</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-3',false)">D) Cannot be determined</button>

</div>

<p><span class="label">Explanation:</span> k is undefined at x = 2, but lim<sub>x→2</sub> k(x) = lim<sub>x→2</sub>(x + 2) = 4 exists. Since the limit exists, this discontinuity is removable.</p>

</div>

<p class="prompt">4. A function f is defined by f(x) = x² + 1 for x ≤ 2, and f(x) = 6 for x > 2. Is f continuous at x = 2?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pd-4',true)">A) No, it has a jump discontinuity</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-4',false)">B) Yes, it is continuous</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-4',false)">C) No, but the discontinuity is removable</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-4',false)">D) Cannot be determined</button>

</div>

<p><span class="label">Explanation:</span> Left-hand limit: lim<sub>x→2⁻</sub>(x² + 1) = 5. Right-hand limit: lim<sub>x→2⁺</sub>(6) = 6. Since these disagree, f has a jump discontinuity, which cannot be removed.</p>

</div>

<p class="prompt">5. A function g is defined by g(x) = x² for x ≠ 3, and g(3) = 20. Is g continuous at x = 3?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pd-5',true)">A) No, but the discontinuity is removable</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-5',false)">B) Yes, it is continuous</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-5',false)">C) No, and the discontinuity is not removable</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-5',false)">D) Cannot be determined</button>

</div>

<p><span class="label">Explanation:</span> lim<sub>x→3</sub> g(x) = lim<sub>x→3</sub> x² = 9 (the limit only cares about values near 3, not the defined value at 3 itself). Since g(3) = 20 ≠ 9, g is not continuous, but redefining g(3) = 9 would remove the discontinuity.</p>

</div>

<p class="prompt">6. What kind of discontinuity does the greatest integer function f(x) = &lfloor;x&rfloor; have at each integer?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pd-6',true)">A) Jump discontinuity</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-6',false)">B) Removable discontinuity</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-6',false)">C) Infinite discontinuity</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pd-6',false)">D) No discontinuity</button>

</div>

<p><span class="label">Explanation:</span> At every integer, the left- and right-hand limits both exist but differ by exactly 1, the signature of a jump discontinuity, which cannot be removed by redefinition.</p>

</div>

<h2 id="evt-ivt">5. The Extreme Value and Intermediate Value Theorems</h2>

<p>The <strong>Extreme Value Theorem</strong>: if f is continuous on the closed interval [a, b], then f attains both a minimum and a maximum value somewhere in that interval. The <strong>Intermediate Value Theorem</strong> (IVT): if f is continuous on [a, b] and M is any number between f(a) and f(b), then there is at least one c in [a, b] where f(c) = M. A key special case: if f(a) and f(b) have opposite signs, f must have a zero somewhere in [a, b].</p>

<p><strong>Worked Example:</strong> Show that f(x) = x³ − 4x − 1 has a root between x = 2 and x = 3.</p>

<p>f is continuous everywhere (it's a polynomial), and f(2) and f(3) have opposite signs. By the Intermediate Value Theorem, there must be a value c in (2, 3) where f(c) = 0.</p>

</div>

<p class="prompt">1. The Extreme Value Theorem guarantees a function attains a minimum and maximum value on an interval, provided the function is…</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pe-1',true)">A) Continuous on a closed interval</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pe-1',false)">B) Continuous on an open interval</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pe-1',false)">C) Differentiable everywhere</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pe-1',false)">D) A polynomial</button>

</div>

<p><span class="label">Explanation:</span> The Extreme Value Theorem specifically requires continuity on a closed interval [a, b]. On an open interval, a function could approach but never actually attain its extreme values.</p>

</div>

<p class="prompt">2. A continuous function f satisfies f(1) = −4 and f(5) = 6. By the Intermediate Value Theorem, must there be a value c in [1, 5] where f(c) = 0?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pe-2',true)">A) Yes, since 0 is between f(1) and f(5)</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pe-2',false)">B) No, the IVT doesn't guarantee this</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pe-2',false)">C) Only if f is a polynomial</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pe-2',false)">D) Cannot be determined without more information</button>

</div>

<p><span class="label">Explanation:</span> Since f is continuous on [1, 5] and 0 lies between f(1) = −4 and f(5) = 6, the IVT guarantees some c in [1, 5] where f(c) = 0.</p>

</div>

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<p><span class="label">Explanation:</span> f(2) = 8 − 4 − 5 = −1. f(3) = 27 − 6 − 5 = 16.</p>

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<p class="prompt">4. Using the previous result, why does the Intermediate Value Theorem guarantee a root of f(x) = x³ − 2x − 5 between x = 2 and x = 3?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'pe-4',true)">A) f is continuous on [2, 3] and f(2), f(3) have opposite signs</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pe-4',false)">B) f is differentiable everywhere</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pe-4',false)">C) f(2) and f(3) are both positive</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'pe-4',false)">D) f has a horizontal asymptote</button>

</div>

<p><span class="label">Explanation:</span> f, as a polynomial, is continuous everywhere. Since f(2) = −1 and f(3) = 16 have opposite signs, the IVT's zero-finding special case guarantees a root somewhere in between.</p>

</div>

<h2 id="mistakes">Common Mistakes to Avoid</h2>

<li><strong>Writing lim f(x) = ∞ as if the limit exists.</strong> This notation describes unbounded behavior, but a limit must be a finite number. Whenever a limit is ±∞, that limit technically does not exist.</li>

<li><strong>Forgetting to check both one-sided limits before concluding a two-sided limit exists.</strong> If the left- and right-hand limits disagree, even if both are finite, the two-sided limit does not exist.</li>

<li><strong>Trying to cancel a factor before confirming it's actually a common factor.</strong> Always factor completely first, then cancel identical factors from numerator and denominator, canceling too early leads to algebra errors.</li>

<li><strong>Confusing a removable discontinuity with a jump discontinuity.</strong> A removable discontinuity has one well-defined limit that just doesn't match (or exist as) the function's value there. A jump discontinuity has two different one-sided limits, there's no single value that could ever fix it.</li>

<li><strong>Applying the Intermediate Value Theorem without first confirming continuity.</strong> The IVT's guarantee depends entirely on the function being continuous over the interval in question, skipping this check can lead to false conclusions for a discontinuous function.</li>

</ul>

<h2 id="faq">Frequently Asked Questions</h2>

<h3>Why does calculus care so much about one-sided limits?</h3>

<p>Many functions worth studying, piecewise definitions, rational functions near a vertical asymptote, step functions, behave differently depending on which direction you approach from. Checking both sides separately is often the only way to correctly determine whether a limit exists at all.</p>

</div>

<h3>How is continuity different from just having a limit at a point?</h3>

<p>A function can have a perfectly well-defined limit at a point without being continuous there, if the function's actual value at that point either doesn't exist or doesn't match the limit. Continuity requires all three conditions to hold: the value exists, the limit exists, and they're equal to each other.</p>

</div>

<h3>Do I need to memorize the exact statement of the Intermediate Value Theorem for the AP exam?</h3>

<p>Yes, and just as importantly, you need to recognize when its hypotheses (continuity on a closed interval) are satisfied before invoking it, simply having a graph that "looks smooth" is not the same as a justified continuity argument.</p>

</div>

<h3>Where can I practice more problems like these?</h3>

<p>The <a href="https://theschoolofmathematics.com/quiz/Limits%26Continuity-Quiz%201">Limits & Continuity quizzes</a> in the AP Calculus AB Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested throughout the course.</p>

</div>

<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/Limits%26Continuity-Quiz%201">Practice Limits & Continuity Free</a>

<a class="cta-btn cta-secondary" href="https://theschoolofmathematics.com/quiz/course/AP-Calculus-AB-QBank">Explore the Full AP Calculus AB Qbank</a>

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