Limit Definition & Notation - Calculus 1

Given $\lim_{x \to 2} \frac{f(x) - 5}{x - 2} = 3$ , which of the following statements must be true?
Answer: $\lim_{x \to 2} f(x) = 5$ When you encounter a limit of the form $\lim_{x \to a} \frac{f(x) - L}{x - a}$ , you're looking at the definition of a derivative. This specific form tells you that $f'(a) = 3$ and reveals crucial information about the function's behavior near $x = 2$ . For this limit to exist and equal 3, the numerator $f(x) - 5$ must approach 0 as $x$ approaches 2. If it didn't, you'd either get an undefined limit (if the numerator approached a non-zero value) or zero (if the numerator approached zero faster than the denominator). Since the limit equals 3, both the numerator and denominator must approach zero together. Therefore, $\lim_{x \to 2} f(x) - 5 = 0$ , which means $\lim_{x \to 2} f(x) = 5$ . This confirms answer choice B. Looking at the wrong answers: A is incorrect because the limit existing doesn't tell us whether $f(2)$ is defined—it could equal 5, be undefined, or equal something else entirely. C is wrong because vertical asymptotes occur when limits approach infinity, not finite values like 3. D confuses the limit of the difference quotient (which is 3) with the limit of the function itself (which is 5). Remember this pattern: when you see $\lim_{x \to a} \frac{f(x) - L}{x - a} = k$ , the key insight is that $\lim_{x \to a} f(x) = L$ . The derivative value $k$ tells you about the slope, but $L$ tells you where the function is heading.
If $\lim_{x \to 3^-} f(x) = 5$ and $f(3)$ is undefined, what can be concluded about $\lim_{x \to 3} f(x)$ ?
Answer: $\lim_{x \to 3} f(x)$ cannot be determined without information about the right-hand limit. The existence of a two-sided limit requires three conditions: the left-hand limit must exist, the right-hand limit must exist, and these two limits must be equal. We are only given that the left-hand limit exists and is 5. We have no information about the right-hand limit, $\lim_{x \to 3^+} f(x)$ . It could be 5, it could be another number, or it could fail to exist. Therefore, we cannot determine the two-sided limit. The fact that $f(3)$ is undefined is irrelevant to the existence of the limit itself.
The Squeeze Theorem is a direct consequence of the $\epsilon-\delta$ definition of a limit. If $g(x) \le f(x) \le h(x)$ and $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$ , how is a suitable $\delta$ for $f(x)$ constructed in the proof?
Answer: For any $\epsilon > 0$ , the $\delta$ for $f(x)$ is the minimum of the $\delta$ 's that work for $g(x)$ and $h(x)$ . For any given $\epsilon > 0$ , since the limits of $g$ and $h$ exist, we can find a $\delta_g$ and a $\delta_h$ . If we choose $\delta = \min(\delta_g, \delta_h)$ , then for any $x$ with $0 < |x-c| < \delta$ , we know that both $|g(x)-L| < \epsilon$ and $|h(x)-L| < \epsilon$ . This means $L-\epsilon < g(x)$ and $h(x) < L+\epsilon$ . Since $g(x) \le f(x) \le h(x)$ , we have $L-\epsilon < f(x) < L+\epsilon$ , which proves that $\lim_{x \to c} f(x) = L$ .
Let $f(x)$ be a function. If for every $M > 0$ , there exists a $\delta > 0$ such that $f(x) > M$ whenever $0 < x - a < \delta$ , which of the following limit notations correctly expresses this property?
Answer: $\lim_{x \to a^+} f(x) = \infty$ The condition $0 < x - a < \delta$ is equivalent to $a < x < a + \delta$ . This means $x$ is approaching $a$ from the right side (from values greater than $a$ ). The condition that for any large number $M$ , we can get $f(x) > M$ by choosing $x$ in this way, is the definition of the right-hand limit of $f(x)$ at $a$ being positive infinity. This is denoted by $\lim_{x \to a^+} f(x) = \infty$ .
Let $f(x) = \lfloor x \rfloor$ be the greatest integer function. Which of the following statements correctly describes $\lim_{x \to 2} f(x)$ using limit notation and justification?
Answer: $\lim_{x \to 2} f(x)$ does not exist because $\lim_{x \to 2^-} f(x) = 1$ and $\lim_{x \to 2^+} f(x) = 2$ . To evaluate the two-sided limit, we must check the one-sided limits. As $x$ approaches 2 from the left (e.g., $x=1.9, 1.99, ...$ ), $f(x) = \lfloor x \rfloor = 1$ . So, $\lim_{x \to 2^-} f(x) = 1$ . As $x$ approaches 2 from the right (e.g., $x=2.1, 2.01, ...$ ), $f(x) = \lfloor x \rfloor = 2$ . So, $\lim_{x \to 2^+} f(x) = 2$ . Since the left-sided limit (1) does not equal the right-sided limit (2), the two-sided limit does not exist.
For the function $f(x) = \sqrt{x}$ , we want to prove $\lim_{x \to 9} \sqrt{x} = 3$ . Given $\epsilon = 0.1$ , we require $|\sqrt{x} - 3| < 0.1$ . Which open interval for $x$ , centered at 9, guarantees this condition?
Answer: $(8.41, 9.61)$ The condition is $|\sqrt{x} - 3| < 0.1$ , which is equivalent to $-0.1 < \sqrt{x} - 3 < 0.1$ . Adding 3 to all parts gives $2.9 < \sqrt{x} < 3.1$ . Squaring all parts gives $(2.9)^2 < x < (3.1)^2$ , which is $8.41 < x < 9.61$ . Any $x$ in this interval will satisfy the condition. The question asks for the interval that guarantees the condition, which is this exact interval derived from the epsilon value.
Which of the following correctly states the formal definition of an infinite limit, $lim_{x->c} f(x) = \infty$ ?
Answer: For every $M > 0$ , there exists a $\delta > 0$ such that if $0 < |x - c| < \delta$ , then $f(x) > M$ . Understanding infinite limits requires grasping how functions behave as they approach vertical asymptotes or points where they "blow up" to infinity. The formal definition captures the idea that we can make the function arbitrarily large by getting sufficiently close to a particular x-value. The correct definition states that for every threshold M > 0 (no matter how large), there exists a δ > 0 such that whenever x is within δ units of c (but not equal to c), the function value f(x) exceeds M. This means we can make f(x) as large as we want by choosing x close enough to c. Option D captures this precisely: "For every M > 0, there exists a δ > 0 such that if $0 < |x - c| < δ$ , then $f(x) > M$ ." Option A reverses the logical structure incorrectly, putting the condition on f(x) in the "if" clause rather than the conclusion. Option B inappropriately uses ε (typically reserved for finite limits) and the condition $|f(x)| > 1/ε$ doesn't properly express that f(x) grows without bound. Option C has the quantifiers backwards—it claims there exists some fixed M that works for all δ, which would mean the function is bounded above, contradicting the infinite limit. Study tip: Remember that infinite limit definitions flip the usual ε-δ structure. Instead of keeping f(x) within ε of a finite value, you're making f(x) exceed any given threshold M. The quantifier order "for every M, there exists δ" is crucial—you must be able to handle any challenge M by finding an appropriate δ.
The velocity $v(t)$ (in m/s) of a skydiver approaches a terminal velocity of 60 m/s as time $t$ increases. Which statement correctly expresses this physical phenomenon using formal limit notation?
Answer: $\lim_{t \to \infty} v(t) = 60$ When you encounter problems about terminal velocity or any situation where a quantity "approaches" a specific value as time goes to infinity, you're dealing with limits at infinity. This is a fundamental concept that connects real-world behavior to formal mathematical language. The phrase "approaches a terminal velocity of 60 m/s as time increases" translates directly to $\lim_{t \to \infty} v(t) = 60$ . This notation captures exactly what's happening: as $t$ gets arbitrarily large (approaches infinity), the velocity function $v(t)$ gets arbitrarily close to 60 m/s. Answer choice D correctly expresses this relationship. Now let's see why the other options miss the mark. Choice A has the formal structure backwards—it claims that for any time period, there exists some error tolerance, when the epsilon-delta definition actually requires that for any given tolerance, we can find a time beyond which the velocity stays within that tolerance. Choice B switches the roles of time and velocity, suggesting that as velocity approaches 60, time goes to infinity—this describes a completely different (and nonsensical) relationship. Choice C uses the notation $v(\infty)$ , but infinity isn't a number you can plug into a function; we can only describe the behavior as the input approaches infinity. Study tip: When you see phrases like "approaches," "tends toward," or "gets closer to" as some variable increases without bound, immediately think "limit as that variable approaches infinity." The mathematical notation should mirror the physical description.
Let $f(x)$ be a function defined as $f(x) = \begin{cases} kx^2 - 3 & \text{if } x < 2 \\ 5 & \text{if } x = 2 \\ x + k & \text{if } x > 2 \end{cases}$ . For what value of the constant $k$ does $lim_{x->2} f(x)$ exist?
Answer: $5/3$ For the two-sided limit $lim_{x->2} f(x)$ to exist, the left-hand limit must equal the right-hand limit. The value of the function at $x=2$ , which is $f(2)=5$ , is irrelevant for the existence of the limit. We calculate the one-sided limits: Left-hand limit: $lim_{x->2^-} f(x) = lim_{x->2^-} (kx^2 - 3) = k(2^2) - 3 = 4k - 3$ . Right-hand limit: $lim_{x->2^+} f(x) = lim_{x->2^+} (x + k) = 2 + k$ . Setting them equal: $4k - 3 = 2 + k$ . Solving for $k$ gives $3k = 5$ , so $k = 5/3$ .
Which of the following statements is the precise definition of $\lim_{x \to \infty} f(x) = L$ ?
Answer: For each $\epsilon > 0$ , there exists a number $N$ such that if $x > N$ , then $|f(x) - L| < \epsilon$ . The definition of a limit at infinity states that we can make $f(x)$ arbitrarily close to $L$ (within $\epsilon$ ) by taking $x$ to be sufficiently large (greater than some number $N$ ). Option A correctly states this. Option B describes the case where the limit is $L$ as $x \to \infty$ and as $x \to -\infty$ . Option C has the quantifiers reversed. Option D has the logical implication reversed.
To prove that $\lim_{x \to 2} x^2 = 4$ using the $\epsilon-\delta$ definition, we start with $|x^2 - 4| < \epsilon$ . To control the $|x+2|$ term that arises from factoring $|x-2||x+2|$ , we make a preliminary restriction that $\delta \le 1$ . Given this restriction, which of the following represents a valid choice for $\delta$ that completes the proof?
Answer: $\delta = \min(1, \epsilon/5)$ We need to bound $|x+2|$ . The restriction $\delta \le 1$ implies $|x-2| < 1$ , which means $-1 < x-2 < 1$ , or $1 < x < 3$ . For $x$ in this interval, $x+2$ is between 3 and 5. Thus, we can bound $|x+2|$ by 5. Now we have $|x^2 - 4| = |x-2||x+2| < |x-2| \cdot 5$ . We want this to be less than $\epsilon$ , so we need $5|x-2| < \epsilon$ , which means $|x-2| < \epsilon/5$ . Since we must satisfy both $|x-2| < 1$ and $|x-2| < \epsilon/5$ , we must choose $\delta = \min(1, \epsilon/5)$ .
Consider the limit $\lim_{x \to c} x^2 = c^2$ . For a fixed $\epsilon > 0$ , the largest possible choice of $\delta$ in the $\epsilon-\delta$ proof depends on both $\epsilon$ and $c$ . As $|c|$ increases, what happens to this largest possible value of $\delta$ ?
Answer: It decreases because the function becomes steeper. The slope of the function $f(x)=x^2$ is $f'(x)=2x$ . As $|c|$ increases, the magnitude of the slope at $x=c$ increases, meaning the function becomes steeper. For a steeper function, a small change in $x$ (the $\delta$ -interval) produces a larger change in $f(x)$ (the $\epsilon$ -interval). Therefore, to keep the change in $f(x)$ within the fixed $\epsilon$ tolerance, we must use a smaller $\delta$ -interval as $|c|$ increases.
Let $f(x)$ and $g(x)$ be functions. If $\lim_{x \to 1} g(x) = 0$ and $\lim_{x \to 1} \frac{f(x)}{g(x)} = 5$ , which of the following statements about $f(x)$ must be true?
Answer: $\lim_{x \to 1} f(x) = 0$ When you encounter a problem involving the limit of a quotient where the denominator approaches zero, you need to think carefully about what this tells you about the numerator. Since $\lim_{x \to 1} \frac{f(x)}{g(x)} = 5$ exists and is finite, and $\lim_{x \to 1} g(x) = 0$ , we can use the limit laws in reverse. If the limit of the quotient exists and the denominator approaches zero, then the numerator must also approach zero. Here's why: if $\lim_{x \to 1} f(x) = L$ where $L \neq 0$ , then as $x \to 1$ , we'd have $\frac{f(x)}{g(x)} \to \frac{L}{0}$ , which would be undefined (infinite). Since the quotient has a finite limit of 5, we must have $\lim_{x \to 1} f(x) = 0$ . Looking at the wrong answers: Choice (A) is incorrect because $\lim_{x \to 1} f(x)$ does exist—it equals 0. Choice (B) incorrectly assumes you can simply multiply the limit of the quotient by the limit of the denominator, but $5 \times 0 = 0$ , not 5. Choice (D) confuses the limit of $f(x)$ as $x$ approaches 1 with the actual value $f(1)$ —these can be different, and we have no information about what $f(1)$ equals. Key takeaway: When a quotient has a finite limit but the denominator approaches zero, the numerator must also approach zero. This creates an indeterminate form $\frac{0}{0}$ that can resolve to any finite value through techniques like L'Hôpital's rule.
You are given that $lim_{x->a} f(x) = L$ . A student is proving that $lim_{x->a} 3f(x) = 3L$ using the $\epsilon-\delta$ definition. The student starts correctly: "For any $\epsilon > 0$ , we need to find a $\delta > 0$ such that if $0 < |x - a| < \delta$ , then $|3f(x) - 3L| < \epsilon$ ." Which of the following is a correct and logical next step in their proof?
Answer: Let $\epsilon' = \epsilon/3$ . From the given limit, we know there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$ , then $|f(x) - L| < \epsilon'$ . When proving limits using the epsilon-delta definition, you need to work backwards from what you want to prove to what you're given. The key insight is connecting the constraint you need ( $|3f(x) - 3L| < \epsilon$ ) to the limit you already know ( $\lim_{x \to a} f(x) = L$ ). The correct approach is option D. Since you need $|3f(x) - 3L| < \epsilon$ , you can factor this as $|3(f(x) - L)| = 3|f(x) - L| < \epsilon$ . This means you need $|f(x) - L| < \epsilon/3$ . By setting $\epsilon' = \epsilon/3$ and using the given limit, you can find a $\delta$ such that $|f(x) - L| < \epsilon'$ , which gives you exactly what you need. Option A gets the inequality backwards – it finds $\delta$ for $|f(x) - L| < \epsilon$ , then multiplies to get $|3f(x) - 3L| < 3\epsilon$ . This gives you more than $\epsilon$ , not less than $\epsilon$ as required. Option B incorrectly sets $\delta = \epsilon/3$ without justification. The relationship between $\epsilon$ and $\delta$ depends on the specific function, not a direct proportionality. Option C sets $\epsilon' = 3\epsilon$ , which would give you $|f(x) - L| < 3\epsilon$ . When you multiply by 3, you'd get $|3f(x) - 3L| < 9\epsilon$ , which is far too large. Study tip: In epsilon-delta proofs involving constants, always work backwards from your target inequality to determine what epsilon value you need for the given limit.
A function $h(x)$ is defined such that $\lim_{x \to 4^-} h(x) = 1$ , $\lim_{x \to 4^+} h(x) = -2$ , and $h(4) = 1$ . Which statement correctly describes the behavior of $h(x)$ at $x=4$ ?
Answer: $\lim_{x \to 4} h(x)$ does not exist because the left-hand and right-hand limits are not equal. For a two-sided limit to exist, the limit from the left and the limit from the right must both exist and be equal. In this case, the left-hand limit is 1 and the right-hand limit is -2. Since $1 \neq -2$ , the two-sided limit $\lim_{x \to 4} h(x)$ does not exist. The value of $h(4)$ is irrelevant to the existence of the limit.
Suppose it is known that $lim_{x->c} f(x) = L$ . According to the $\epsilon-\delta$ definition, for any chosen $\epsilon > 0$ , there exists a $\delta > 0$ that satisfies the required conditions. If $\delta_1$ is a specific value of $\delta$ that is shown to work for a given $\epsilon_1$ , which of the following statements must also be true?
Answer: For the same $\epsilon_1$ , any $\delta_2$ such that $0 < \delta_2 < \delta_1$ is also a valid choice for $\delta$ . The $\epsilon-\delta$ definition states that if $0 < |x - c| < \delta_1$ , then $|f(x) - L| < \epsilon_1$ . If we choose a $\delta_2$ such that $0 < \delta_2 < \delta_1$ , then the set of $x$ values satisfying $0 < |x - c| < \delta_2$ is a subset of the $x$ values satisfying $0 < |x - c| < \delta_1$ . Therefore, the condition $|f(x) - L| < \epsilon_1$ must also hold for this smaller set of $x$ values, making $\delta_2$ a valid choice.
Let $f(x)$ be defined as $f(x) = 2x$ if $x$ is rational, and $f(x) = x+3$ if $x$ is irrational. Using the conceptual basis of the $\epsilon-\delta$ definition, what is $\lim_{x \to 3} f(x)$ ?
Answer: 6 For the limit to exist, the function values must approach a single number $L$ as $x$ approaches 3, regardless of whether $x$ is rational or irrational. As $x \to 3$ along rational numbers, $f(x) = 2x \to 2(3) = 6$ . As $x \to 3$ along irrational numbers, $f(x) = x+3 \to 3+3 = 6$ . Since both paths approach the same value, 6, the limit is 6. At any other point $c$ , where $2c \neq c+3$ , the limit would not exist.
Consider the limit $\lim_{x \to 3} (4x - 5) = 7$ . If $\epsilon = 0.01$ , what is the largest value of $\delta$ such that if $0 < |x - 3| < \delta$ , then $|(4x - 5) - 7| < 0.01$ ?
Answer: 0.0025 We start with the inequality $|(4x - 5) - 7| < 0.01$ . Simplifying the expression inside the absolute value gives $|4x - 12| < 0.01$ . We can factor out a 4: $4|x - 3| < 0.01$ . To find the condition on $|x - 3|$ , we divide both sides by 4: $|x - 3| < 0.01 / 4$ , which simplifies to $|x - 3| < 0.0025$ . Therefore, the largest possible value for $\delta$ is 0.0025.
Consider the function $f(x) = \cos(\frac{\pi}{x})$ . Which statement best describes the limit of $f(x)$ as $x$ approaches 0?
Answer: $lim_{x->0} f(x)$ does not exist because the function oscillates between -1 and 1. When analyzing limits involving trigonometric functions, you need to carefully examine what happens to both the function's argument and its behavior as you approach the point in question. For $f(x) = \cos(\frac{\pi}{x})$ , let's see what happens as $x \to 0$ . As $x$ gets closer to 0, the expression $\frac{\pi}{x}$ becomes very large in absolute value. For positive values approaching 0, $\frac{\pi}{x} \to +\infty$ , and for negative values, $\frac{\pi}{x} \to -\infty$ . Since cosine oscillates between -1 and 1 for all real inputs, and $\frac{\pi}{x}$ takes on arbitrarily large values as $x \to 0$ , the function $\cos(\frac{\pi}{x})$ oscillates rapidly between -1 and 1 near $x = 0$ . No matter how small a neighborhood around 0 you choose, the function will complete infinitely many oscillations within that interval. This means the limit does not exist because the function values don't approach a single number. Option A is incorrect because a limit can exist even when the function is undefined at that point—limits describe nearby behavior, not the value at the point itself. Option B is wrong because cosine is always bounded between -1 and 1, so the function never becomes unbounded. Option D misapplies limit reasoning—while it's true that as $x \to 0^+$ , we might think $\frac{\pi}{x} \to 0$ , this is backwards since $\frac{\pi}{x}$ actually grows without bound. Study tip: When evaluating limits of composite functions, always track what happens to the inner function first, then consider how the outer function behaves with those inputs.
Let $f(x)$ be a function and $c$ be a real number within an open interval in the domain of $f$ . Which of the following scenarios is impossible?
Answer: $f(x)$ is continuous at $x=c$ , but $\lim_{x \to c} f(x)$ does not exist. The definition of continuity at a point $x=c$ requires three conditions to be met: 1) $f(c)$ must be defined, 2) $\lim_{x \to c} f(x)$ must exist, and 3) $\lim_{x \to c} f(x) = f(c)$ . Scenario (D) states that the function is continuous at $x=c$ and that the limit does not exist. This is a direct contradiction of the definition of continuity. Scenarios A, B, and C all describe possible types of discontinuities (removable and jump discontinuities).

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