Limits via Algebraic Properties - Calculus 1

1. Given $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$, where $L$ and $M$ are finite real numbers. Under which condition can we NOT conclude that $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M}$?
   Answer: $M = 0$ The quotient rule for limits states that if $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$ both exist as finite numbers, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M}$, provided that $M \neq 0$. If $M=0$, the quotient rule cannot be applied, and the limit may be infinite or may require further analysis if it is of the indeterminate form $0/0$. Therefore, the conclusion is not guaranteed when $M=0$. The values of $f(c)$ and $g(c)$ are irrelevant to the value of the limit as $x$ approaches $c$.
2. Calculate the following limit: $$\lim_{x \to 3} \frac{\sqrt{x+1} - 2}{x^2 - 9}$$
   Answer: $\frac{1}{24}$ This limit is of the indeterminate form $\frac{0}{0}$. To resolve this, we multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x+1} + 2$: $$\lim_{x \to 3} \frac{\sqrt{x+1} - 2}{x^2 - 9} \cdot \frac{\sqrt{x+1} + 2}{\sqrt{x+1} + 2} = \lim_{x \to 3} \frac{(x+1) - 4}{(x-3)(x+3)(\sqrt{x+1} + 2)}$$ Simplify the numerator: $$\lim_{x \to 3} \frac{x - 3}{(x-3)(x+3)(\sqrt{x+1} + 2)}$$ Cancel the $(x-3)$ terms: $$\lim_{x \to 3} \frac{1}{(x+3)(\sqrt{x+1} + 2)}$$ Substitute $x = 3$: $$\frac{1}{(3+3)(\sqrt{3+1} + 2)} = \frac{1}{6(\sqrt{4} + 2)} = \frac{1}{6(2+2)} = \frac{1}{24}$$ B: This common error occurs if one forgets to include the conjugate term in the denominator after multiplying. C: This results from incorrectly evaluating the final expression or guessing. D: This is a common guess for limits that initially appear to have a zero in the denominator.
3. Evaluate the limit: $\lim_{x \to 0} \frac{(x+2)^{-1} - 2^{-1}}{x}$
   Answer: $-\frac{1}{4}$ The limit is of the indeterminate form $\frac{0}{0}$. First, rewrite the negative exponents as fractions: $\lim_{x \to 0} \frac{\frac{1}{x+2} - \frac{1}{2}}{x}$ Combine the fractions in the numerator by finding a common denominator, which is $2(x+2)$: $\lim_{x \to 0} \frac{\frac{2 - (x+2)}{2(x+2)}}{x} = \lim_{x \to 0} \frac{\frac{-x}{2(x+2)}}{x}$ Now, simplify the complex fraction: $\lim_{x \to 0} \frac{-x}{2x(x+2)}$ Cancel the $x$ terms: $\lim_{x \to 0} \frac{-1}{2(x+2)}$ Substitute $x=0$: $ \frac{-1}{2(0+2)} = -\frac{1}{4} $ A: A common guess for limits that initially appear to have a zero in the denominator. B: This can result from forgetting the '2' from the $\frac{1}{2}$ term in the common denominator. C: This choice indicates a sign error.
4. Evaluate the limit, which represents the derivative of $f(x) = \sqrt[3]{x}$ at $x=8$: $$\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x-8}$$
   Answer: $\frac{1}{12}$ The limit is of the indeterminate form $\frac{0}{0}$. We can resolve this by recognizing the denominator as a difference of cubes: $x-8 = (\sqrt[3]{x})^3 - 2^3$. Using the factorization $a^3 - b^3 = (a-b)(a^2+ab+b^2)$, we get: $$x-8 = (\sqrt[3]{x}-2)((\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4)$$ Substitute this into the limit expression: $$\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{(\sqrt[3]{x}-2)(x^{2/3} + 2x^{1/3} + 4)}$$ Cancel the $(\sqrt[3]{x}-2)$ term: $$\lim_{x \to 8} \frac{1}{x^{2/3} + 2x^{1/3} + 4}$$ Now substitute $x=8$: $$\frac{1}{8^{2/3} + 2(8^{1/3}) + 4} = \frac{1}{(2)^2 + 2(2) + 4} = \frac{1}{4+4+4} = \frac{1}{12}$$ B: This can result from an incomplete factorization of the difference of cubes, perhaps missing the middle term $ab$. C: This results from using only one term of the denominator polynomial after substitution. D: This is an incorrect guess for a $\frac{0}{0}$ limit.
5. Evaluate the limit: $$\lim_{x \to 2} \frac{\sqrt{x+7} - 3}{x-5}$$
   Answer: 0 This question is a test of whether students check for indeterminate forms before applying more complex methods. The first step for any limit is direct substitution. Substitute $x = 2$ into the expression: Numerator: $\sqrt{2+7} - 3 = \sqrt{9} - 3 = 3 - 3 = 0$. Denominator: $2 - 5 = -3$. The limit is $\frac{0}{-3} = 0$. B: This is the answer to a different, very common limit problem, $\lim_{x \to 2} \frac{\sqrt{x+7}-3}{x-2}$. Students who see the radical and assume they must use the conjugate method without first substituting may mistakenly arrive at this answer. C: This might be chosen if a student makes an error during substitution and believes the denominator is zero. D: This is the value of the denominator, not the limit itself.
6. Let $f(x) = \frac{2x^2 + ax + b}{x-3}$. If $\lim_{x \to 3} f(x) = 10$, what is the value of the sum $a+b$?
   Answer: -14 For the limit of the rational function to exist as $x \to 3$, while the denominator approaches 0, the numerator must also approach 0. This creates an indeterminate form that can be simplified. First, set the numerator to 0 at $x=3$: $2(3)^2 + a(3) + b = 0 \implies 18 + 3a + b = 0$ This means $(x-3)$ must be a factor of the numerator. We can use this fact to find the limit in terms of $a$: $\lim_{x \to 3} \frac{2x^2 + ax + b}{x-3} = 10$ Using L'Hôpital's Rule (or by factoring), we differentiate the numerator and denominator: $\lim_{x \to 3} \frac{4x+a}{1} = 10$ Substitute $x=3$: $4(3) + a = 10 \implies 12 + a = 10 \implies a = -2$ Now substitute $a=-2$ back into the first equation: $18 + 3(-2) + b = 0 \implies 18 - 6 + b = 0 \implies 12 + b = 0 \implies b = -12$ The question asks for $a+b$, which is $-2 + (-12) = -14$. A: This is the value of $a$, not $a+b$. B: This is the value of $b$, not $a+b$. D: This answer may result from a sign error when solving for $b$.
7. For a constant $a > 0$, evaluate the limit: $$\lim_{h \to 0} \frac{\sqrt{a^2+h} - a}{h}$$
   Answer: $\frac{1}{2a}$ This limit is of the indeterminate form $\frac{0}{0}$. We multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{a^2+h} + a$: $$\lim_{h \to 0} \frac{\sqrt{a^2+h} - a}{h} \cdot \frac{\sqrt{a^2+h} + a}{\sqrt{a^2+h} + a} = \lim_{h \to 0} \frac{(a^2+h) - a^2}{h(\sqrt{a^2+h} + a)}$$ Simplify the numerator: $$\lim_{h \to 0} \frac{h}{h(\sqrt{a^2+h} + a)}$$ Cancel the $h$ terms: $$\lim_{h \to 0} \frac{1}{\sqrt{a^2+h} + a}$$ Substitute $h=0$: $$\frac{1}{\sqrt{a^2} + a} = \frac{1}{a + a} = \frac{1}{2a}$$ (Note: This is the definition of the derivative of $f(x)=\sqrt{x}$ at $x=a^2$). A: An incorrect guess. B: This common mistake arises from forgetting the second $a$ in the denominator after taking the limit (i.e., $\sqrt{a^2}=a$, but then forgetting to add the other $a$). D: This might result from confusion with the power rule for derivatives in a different form.
8. Evaluate the limit: $\lim_{x \to \sqrt{2}} \frac{x^4 - 4}{x^2 - 2}$
   Answer: 4 Direct substitution of $x = \sqrt{2}$ yields the indeterminate form $\frac{0}{0}$. The numerator $x^4-4$ can be factored as a difference of squares. $x^4 - 4 = (x^2)^2 - 2^2 = (x^2 - 2)(x^2 + 2)$ Substitute this factorization into the limit: $\lim_{x \to \sqrt{2}} \frac{(x^2 - 2)(x^2 + 2)}{x^2 - 2}$ Cancel the common factor $(x^2 - 2)$: $\lim_{x \to \sqrt{2}} (x^2 + 2)$ Now substitute $x = \sqrt{2}$: $ (\sqrt{2})^2 + 2 = 2 + 2 = 4 $ A: A common incorrect guess for $\frac{0}{0}$ limits. B: This might result from incorrectly evaluating $(\sqrt{2})^2$ as $\sqrt{2}$ or some other arithmetic error. D: This might result from incorrectly evaluating the expression as $x^4+4$ and substituting, giving $4+4=8$.
9. For what value of the constant $k$ does the following limit exist? $\lim_{x \to -1} \frac{x^2 - kx - 5}{x+1}$
   Answer: 4 For a limit of a rational function to exist where the denominator approaches 0, the numerator must also approach 0. This creates a $\frac{0}{0}$ indeterminate form which may resolve to a finite value. Set the numerator equal to 0 at $x = -1$: $(-1)^2 - k(-1) - 5 = 0$ $1 + k - 5 = 0$ $k - 4 = 0$ $k = 4$ A: This is the value of the limit itself if $k=4$. $\lim_{x \to -1} \frac{x^2 - 4x - 5}{x+1} = \lim_{x \to -1} \frac{(x-5)(x+1)}{x+1} = -1-5 = -6$. Students might solve for $k$ and then solve for the limit, choosing the limit's value instead of $k$'s. B: This results from a sign error in solving for $k$, such as $1-k-5=0$. D: This may result from misplacing a term in the setup equation.
10. Evaluate the limit: $$\lim_{x \to -2} \frac{x^3 + 8}{x^2 - 4}$$
    Answer: -3 The limit is of the indeterminate form $\frac{0}{0}$. We can factor the numerator as a sum of cubes and the denominator as a difference of squares. $$\lim_{x \to -2} \frac{(x+2)(x^2 - 2x + 4)}{(x-2)(x+2)}$$ Cancel the $(x+2)$ terms: $$\lim_{x \to -2} \frac{x^2 - 2x + 4}{x-2}$$ Now substitute $x = -2$: $$\frac{(-2)^2 - 2(-2) + 4}{-2 - 2} = \frac{4 + 4 + 4}{-4} = \frac{12}{-4} = -3$$ A: This is a common guess for limits with a zero in the denominator before simplification. C: This results from an incorrect factorization of the sum of cubes, such as $(x+2)(x^2+2x+4)$. D: This results from sign errors during substitution, such as $\frac{4+4+4}{-2+2}$ which is undefined, or $\frac{12}{2}$ from sign errors in both numerator and denominator.
11. Let $f(x) = \begin{cases} x^2+3 & \text{if } x < 2 \\ 2x+3 & \text{if } x \ge 2 \end{cases}$. Find $\lim_{x \to 1} \frac{f(x) - f(1)}{x-1}$.
    Answer: 2 The limit expression is the definition of the derivative of $f(x)$ at $x=1$. To evaluate it, we must first determine which piece of the function $f(x)$ is relevant for $x$ near 1. Since 1 is less than 2, we use the first piece, $f(x) = x^2+3$. First, find $f(1)$: $$f(1) = 1^2 + 3 = 4$$ Now substitute this into the limit expression: $$\lim_{x \to 1} \frac{(x^2+3) - 4}{x-1} = \lim_{x \to 1} \frac{x^2 - 1}{x-1}$$ This is a $\frac{0}{0}$ indeterminate form. Factor the numerator: $$\lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 1+1=2$$ A: This is the value of $f(1)$, not the value of the limit. C: This is the limit of the second piece of the function as x approaches 2, i.e. $f(2)$ does not equal the limit of the first piece, but this is irrelevant for the limit at x=1. D: This might be chosen by students who are confused by the piecewise definition and assume a discontinuity at $x=1$, but the function is continuous and differentiable at $x=1$.
12. Let $f(x) = x-2$ and $g(x) = \frac{1}{x-2}$. What is $\lim_{x \to 2} (f(x) \cdot g(x))$?
    Answer: $1$ When you encounter limits involving products where one factor approaches zero and another approaches infinity, you need to simplify the expression first rather than trying to evaluate each factor separately. Let's compute the product $f(x) \cdot g(x) = (x-2) \cdot \frac{1}{x-2}$. For any value of $x$ where $x \neq 2$, we can simplify this expression by canceling the $(x-2)$ terms: $(x-2) \cdot \frac{1}{x-2} = 1$ Since the simplified expression equals 1 for all $x \neq 2$, we have $\lim_{x \to 2} (f(x) \cdot g(x)) = \lim_{x \to 2} 1 = 1$ This makes D correct. A is wrong because even though $g(x) = \frac{1}{x-2}$ approaches infinity as $x$ approaches 2, the product doesn't blow up due to the cancellation with the $(x-2)$ factor from $f(x)$. B is wrong because even though $f(x) = x-2$ approaches zero as $x$ approaches 2, this creates a $0 \cdot \infty$ indeterminate form, not a limit of 0. You must resolve the indeterminate form first. C is wrong because the limit does exist after simplification. The fact that we have an indeterminate form initially doesn't mean the limit doesn't exist—it means we need to work harder to find it. Study tip: When you see indeterminate forms like $0 \cdot \infty$, always try algebraic simplification first. Look for common factors that cancel before attempting more advanced techniques like L'Hôpital's rule.
13. Let $f(x)$ be a function such that for all $x \ne 1$ in an open interval containing 1, the inequality $1 - (x-1)^2 \le f(x) \le \cos(\pi(x-1))$ holds. What is $\lim_{x \to 1} f(x)$?
    Answer: 1 This problem requires the use of the Squeeze Theorem. We must evaluate the limits of the lower and upper bounding functions as $x \to 1$. Limit of the lower bound: $$\lim_{x \to 1} (1 - (x-1)^2) = 1 - (1-1)^2 = 1 - 0 = 1$$ Limit of the upper bound: $$\lim_{x \to 1} \cos(\pi(x-1)) = \cos(\pi(1-1)) = \cos(0) = 1$$ Since $f(x)$ is squeezed between two functions that both approach 1 as $x \to 1$, by the Squeeze Theorem, $\lim_{x \to 1} f(x)$ must also be 1. A: This might be chosen if the first limit is incorrectly evaluated to 0. C: A student might choose this if they are unable to evaluate one of the limits. D: This is the value of $\cos(\pi)$, which would result from incorrectly evaluating the argument of cosine as $\pi$ instead of 0.
14. Suppose $\lim_{x \to 2} g(x) = -3$ and $f(x) = \frac{x^2 - 9}{x+3}$. The function $f$ has a removable discontinuity at $x=-3$. What is $\lim_{x \to 2} f(g(x))$?
    Answer: -6 We can use the property for limits of composite functions: if $\lim_{u \to L} f(u)$ exists, then $\lim_{x \to c} f(g(x)) = \lim_{u \to L} f(u)$ where $L = \lim_{x \to c} g(x)$. First, find the inner limit: $$L = \lim_{x \to 2} g(x) = -3$$ Now we need $\lim_{u \to -3} f(u)$: $$\lim_{u \to -3} f(u) = \lim_{u \to -3} \frac{u^2 - 9}{u+3}$$ This is an indeterminate form $\frac{0}{0}$. Factor the numerator: $$\lim_{u \to -3} \frac{(u-3)(u+3)}{u+3} = \lim_{u \to -3} (u-3) = -3 - 3 = -6$$ B: This is the value of $f(2) = \frac{4-9}{5} = -1$, which incorrectly evaluates the outer function before considering the inner limit. C: This might be chosen because $f(-3)$ is technically undefined, but the limit exists. D: This results from a sign error when evaluating the final limit, $-3+3=0$.
15. Evaluate the one-sided limit: $$\lim_{x \to 1^-} \frac{x^2 - 1}{|x-1|}$$
    Answer: -2 Since the limit approaches 1 from the left ($x \to 1^-$), values of $x$ are slightly less than 1. This means the term $(x-1)$ is negative. Therefore, the absolute value function is defined as $|x-1| = -(x-1)$. $$\lim_{x \to 1^-} \frac{x^2 - 1}{-(x-1)}$$ Factor the numerator: $$\lim_{x \to 1^-} \frac{(x-1)(x+1)}{-(x-1)}$$ Cancel the $(x-1)$ terms: $$\lim_{x \to 1^-} -(x+1)$$ Substitute $x=1$: $$-(1+1) = -2$$ A: This is the value of the limit from the right, $\lim_{x \to 1^+}$, where $|x-1| = x-1$. C: This would be the correct answer for the two-sided limit, $\lim_{x \to 1}$, because the left-hand and right-hand limits are not equal. D: This might be chosen if the student incorrectly cancels terms, leading to an incorrect result.
16. Let $f$ be a function such that $-x^2 + 6x - 7 \le f(x) \le x^2 - 4x + 5$ for all $x$. Find $\lim_{x \to 2} (x \cdot f(x))$.
    Answer: $2$ When you encounter a limit problem involving inequalities that bound a function, think about the Squeeze Theorem (also called the Sandwich Theorem). This powerful tool lets you find limits even when you don't know the exact form of the function. First, let's examine what happens to the bounding functions as $x$ approaches 2. For the lower bound: $-x^2 + 6x - 7$, when $x = 2$, we get $-4 + 12 - 7 = 1$. For the upper bound: $x^2 - 4x + 5$, when $x = 2$, we get $4 - 8 + 5 = 1$. Since both bounds approach the same value, $\lim_{x \to 2} f(x) = 1$ by the Squeeze Theorem. Now we can find $\lim_{x \to 2} (x \cdot f(x))$ using the product rule for limits: $\lim_{x \to 2} x \cdot \lim_{x \to 2} f(x) = 2 \cdot 1 = 2$. Looking at the wrong answers: (A) 0 might tempt you if you incorrectly calculated one of the bounds or forgot to multiply by the limit of $x$. (B) 1 is a common error where you find $\lim_{x \to 2} f(x) = 1$ correctly but forget to multiply by 2. (C) Does Not Exist would only be correct if the bounds approached different values, which they don't. Study tip: When you see a function squeezed between two expressions, immediately check if those bounds have the same limit at the point of interest. If they do, you can apply the Squeeze Theorem confidently, even without knowing the exact function.
17. Find the value of the limit: $$\lim_{x \to 0} \frac{x^2}{1 - \sqrt{1-x^2}}$$
    Answer: 2 Direct substitution gives the indeterminate form $\frac{0}{0}$. To solve this, we can multiply the numerator and denominator by the conjugate of the denominator, which is $1 + \sqrt{1-x^2}$. $$\lim_{x \to 0} \frac{x^2}{1 - \sqrt{1-x^2}} \cdot \frac{1 + \sqrt{1-x^2}}{1 + \sqrt{1-x^2}}$$ $$= \lim_{x \to 0} \frac{x^2(1 + \sqrt{1-x^2})}{1 - (1-x^2)}$$ Simplify the denominator: $$= \lim_{x \to 0} \frac{x^2(1 + \sqrt{1-x^2})}{1 - 1 + x^2} = \lim_{x \to 0} \frac{x^2(1 + \sqrt{1-x^2})}{x^2}$$ Cancel the $x^2$ terms: $$= \lim_{x \to 0} (1 + \sqrt{1-x^2})$$ Now substitute $x=0$: $$1 + \sqrt{1-0} = 1 + 1 = 2$$ A: This is the result of direct substitution into the numerator. B: This is the reciprocal of the correct answer, a common mistake. D: This might be chosen by a student who is unable to simplify the expression.
18. Suppose $\lim_{x \to a} f(x) = 4$ and $\lim_{x \to a} g(x) = -2$. What is the value of $\lim_{x \to a} \frac{2f(x) - (g(x))^2}{\sqrt{f(x)}}$?
    Answer: $2$ When you encounter limit problems with multiple functions, the key is applying limit laws systematically. These laws allow you to break down complex expressions into simpler parts when the individual limits exist. Since $\lim_{x \to a} f(x) = 4$ and $\lim_{x \to a} g(x) = -2$, you can find the limit of the given expression by evaluating each component. First, find the numerator: $\lim_{x \to a} [2f(x) - (g(x))^2] = 2 \cdot 4 - (-2)^2 = 8 - 4 = 4$. Next, find the denominator: $\lim_{x \to a} \sqrt{f(x)} = \sqrt{4} = 2$. Therefore, $\lim_{x \to a} \frac{2f(x) - (g(x))^2}{\sqrt{f(x)}} = \frac{4}{2} = 2$. Let's examine why the other answers are incorrect. Choice A) $5$ likely comes from miscalculating the numerator as $2(4) - (-2) = 10$ instead of properly squaring $g(x)$. Choice C) $6$ might result from errors like computing $2(4) - 2 = 6$ (forgetting to square the negative) or making arithmetic mistakes in the final division. Choice D) $4$ represents stopping at the numerator calculation without dividing by the denominator, essentially forgetting the fraction structure entirely. The correct answer is B) $2$. Remember this pattern: when working with limits of composite expressions, apply limit laws step-by-step to each operation (addition, multiplication, division, powers, roots). Always double-check your arithmetic, especially with negative numbers and exponents, as these are common sources of errors in limit calculations.
19. Evaluate the limit: $$\lim_{x \to 1} \frac{x - 1}{x^{1/3} - 1}$$
    Answer: 3 This limit is of the form $\frac{0}{0}$. A helpful technique is to make a substitution. Let $u = x^{1/3}$. As $x \to 1$, $u \to 1^{1/3} = 1$. Also, $x = u^3$. Substituting these into the limit gives: $$\lim_{u \to 1} \frac{u^3 - 1}{u - 1}$$ This is a standard limit that can be solved by factoring the numerator as a difference of cubes: $$\lim_{u \to 1} \frac{(u-1)(u^2+u+1)}{u-1}$$ Cancel the $(u-1)$ terms: $$\lim_{u \to 1} (u^2+u+1) = 1^2 + 1 + 1 = 3$$ A: This is the reciprocal of the correct answer, which could result from inverting the expression. B: A common incorrect guess for $\frac{0}{0}$ limits. D: This might result from an algebraic error after cancellation.
20. Given $\lim_{x \to 5} f(x) = 10$ and $\lim_{x \to 5} g(x) = 0$. Additionally, it is known that $g(x) > 0$ for all $x$ in the interval $(4, 6)$. Which of the following must be true about $\lim_{x \to 5} \frac{f(x)}{g(x)}$?
    Answer: The limit is $+\infty$. When you encounter limits involving division where the denominator approaches zero, you're dealing with a potential infinite limit situation. The key is analyzing the signs and behaviors of both the numerator and denominator. Here, $\lim_{x \to 5} f(x) = 10$ means $f(x)$ approaches the positive value 10. Meanwhile, $\lim_{x \to 5} g(x) = 0$ with $g(x) > 0$ in the interval $(4,6)$ means $g(x)$ approaches 0 from the positive side (since $g(x)$ is always positive near $x = 5$). When you divide a number approaching a positive constant by a number approaching zero from the positive side, the quotient grows without bound toward positive infinity. Think of it as $\frac{10}{\text{very small positive number}}$ - this gives you a very large positive result. As $g(x)$ gets closer to zero while staying positive, $\frac{f(x)}{g(x)}$ becomes arbitrarily large and positive. Choice A is wrong because we get positive infinity, not negative infinity, since both $f(x)$ and $g(x)$ approach positive values. Choice B incorrectly suggests the limit is finite - but dividing by values approaching zero typically creates infinite limits, not zero. Choice D is incorrect because we have sufficient information: knowing the signs and limiting behaviors of both functions completely determines the behavior of their quotient. Remember: when the numerator approaches a non-zero constant and the denominator approaches zero, always check the sign of the denominator near the limit point. The sign determines whether you get $+\infty$ or $-\infty$.

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