Calculus 1 Quiz: Limit Definition And Notation

What this quiz covers

This quiz focuses on Limit Definition And Notation, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\lim_{x \to 3} f(x) = 5$.
2. $\lim_{x \to 3} f(x)$ does not exist because $f(3)$ is undefined.
3. $\lim_{x \to 3} f(x)$ cannot be determined without information about the right-hand limit. (correct answer)
4. $\lim_{x \to 3} f(x)$ must be different from 5.

Question 2

1. 0.01
2. 0.04
3. 0.0025 (correct answer)
4. 0.005

Question 3

1. First, find a $\delta$ such that $|f(x) - L| < \epsilon$. Then multiplying by 3 gives $|3f(x) - 3L| < 3\epsilon$, which completes the proof.
2. Let $\delta = \epsilon/3$. Then for any $x$ such that $0 < |x - a| < \epsilon/3$, we can conclude that $|3f(x) - 3L| < \epsilon$.
3. Let $\epsilon' = 3\epsilon$. From the given limit, we know there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \epsilon'$.
4. Let $\epsilon' = \epsilon/3$. From the given limit, we know there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \epsilon'$. (correct answer)

Question 4

1. The limit is proven to be 5, i.e., $lim_{x->2} f(x) = 5$.
2. For the choice $\epsilon = 0.1$, any $\delta$ such that $0 < \delta \le 0.02$ is a valid choice. (correct answer)
3. For any new choice of $\epsilon < 0.1$, the corresponding value of $\delta$ must be less than 0.02.
4. The function $f(x)$ must be defined at $x=2$ and satisfy $f(2) = 5$.

Question 5

1. $5/3$ (correct answer)
2. $2$
3. $3$
4. $-1/3$

Question 6

1. For the same $\epsilon_1$, any $\delta_2$ such that $0 < \delta_2 < \delta_1$ is also a valid choice for $\delta$. (correct answer)
2. For the same $\epsilon_1$, any $\delta_2$ such that $\delta_2 > \delta_1$ is also a valid choice for $\delta$.
3. If $0 < \epsilon_2 < \epsilon_1$, then any valid $\delta$ for $\epsilon_2$ must be less than $\delta_1$.
4. The value of $\delta_1$ is uniquely determined by the function $f(x)$ and the value of $\epsilon_1$.

Question 7

1. The function $f(x)$ must be a constant function.
2. The function $f(x)$ must be bounded for all $x \neq c$. (correct answer)
3. The limit of $f(x)$ as $x$ approaches $c$ must be $L$.
4. The function $f(x)$ must be continuous for all $x \neq c$.

Question 8

1. The statement provides evidence that $\lim_{x \to 3} f(x)$ might be 7, but does not prove it. (correct answer)
2. The statement is sufficient to prove that $\lim_{x \to 3} f(x) = 7$.
3. The statement is sufficient to prove that $f(3) = 7$.
4. The statement implies that for any $\epsilon > 0$, the choice $\delta = 0.5$ will satisfy the limit definition.

Question 9

1. $\delta = 0.32$
2. $\delta = 0.08$
3. $\delta = 0.02$ (correct answer)
4. $\delta = 0.01$

Question 10

1. $\lim_{t \to 12} C(t) = \infty$
2. $\lim_{t \to 12^-} C(t) = \infty$ (correct answer)
3. $\lim_{t \to 12^+} C(t) = -\infty$
4. $\lim_{t \to \infty} C(t) = -1$

Question 11

1. $L = M$, and in this case, $K=L=M$. (correct answer)
2. $f(a)$ is defined, and in this case, $K=f(a)$.
3. $L$ and $M$ are both defined, and in this case, $K$ is their average, $(L+M)/2$.
4. $f(a) = L$ and $f(a) = M$, ensuring the function connects with its one-sided limits.

Question 12

1. If $0 < |x - 3| < 0.05$, then $7.9 < f(x) < 8.1$. (correct answer)
2. If $0 < |x - 3| < 0.1$, then $7.95 < f(x) < 8.05$.
3. If $7.9 < f(x) < 8.1$, then $0 < |x - 3| < 0.05$.
4. The value of $f(3)$ must be between 7.9 and 8.1.

Question 13

1. $\lim_{x \to a} f(x) = \infty$
2. $\lim_{x \to \infty} f(x) = a$
3. $\lim_{x \to a^-} f(x) = \infty$
4. $\lim_{x \to a^+} f(x) = \infty$ (correct answer)

Question 14

1. 6 (correct answer)
2. Does not exist because the function's definition switches between two formulas.
3. Does not exist because for any $x$ near 3, there are both rational and irrational numbers.
4. 9

Question 15

1. It increases because the function values are larger.
2. It remains constant because $\epsilon$ is fixed.
3. It decreases because the function becomes steeper. (correct answer)
4. It is independent of $c$ for any quadratic function.

Question 16

1. For every $\epsilon > 0$ and for every $N$, there is some $x > N$ for which $|f(x) - L| \ge \epsilon$.
2. There exists an $\epsilon > 0$ such that for every $N$, there is some $x > N$ for which $|f(x) - L| \ge \epsilon$. (correct answer)
3. There exists an $\epsilon > 0$ and there exists an $N$ such that for all $x > N$, $|f(x) - L| \ge \epsilon$.
4. For every $N$, there exists an $\epsilon > 0$ such that for all $x > N$, $|f(x) - L| \ge \epsilon$.

Question 17

1. A $\delta$ is chosen for $f(x)$ first, which then determines the valid $\epsilon$ for $g(x)$ and $h(x)$.
2. The $\delta$ for $f(x)$ is the average of the $\delta$'s that work for $g(x)$ and $h(x)$ for a given $\epsilon$.
3. For any $\epsilon > 0$, the $\delta$ for $f(x)$ is the minimum of the $\delta$'s that work for $g(x)$ and $h(x)$. (correct answer)
4. The Squeeze Theorem does not require the $\epsilon-\delta$ definition; it is an independent axiom.

Question 18

1. $(8.41, 9.61)$ (correct answer)
2. $(8.9, 9.1)$
3. $(8.4, 9.6)$
4. $(8.1, 9.9)$

Question 19

1. For any given tolerance $\epsilon > 0$, the values of $f(x)$ can be made to lie within $\epsilon$ of $L$ by choosing $x$ sufficiently close to $c$.
2. The function $f(x)$ must be defined at the point $x = c$. (correct answer)
3. The value $L$ is a finite real number.
4. As $x$ gets arbitrarily close to $c$, $f(x)$ gets arbitrarily close to $L$.

Question 20

1. $\delta = \epsilon/5$
2. $\delta = \min(1, \epsilon/4)$
3. $\delta = \min(1, \epsilon/3)$
4. $\delta = \min(1, \epsilon/5)$ (correct answer)