Limits via Algebraic Manipulation - Calculus 1

If the limit $L = \lim_{x\to 2} \frac{x^2 + kx - 10}{x^2 - 4}$ exists and is finite, what is the value of $L$ ?
Answer: $7/4$ For the limit of the rational function to be finite as $x \to 2$ , the denominator approaches $2^2 - 4 = 0$ . Therefore, the numerator must also approach $0$ to avoid an infinite limit. We set the numerator to $0$ at $x=2$ : $2^2 + k(2) - 10 = 0$ , which gives $4 + 2k - 10 = 0$ , so $2k = 6$ , and $k=3$ . Now, substitute $k=3$ into the limit: $L = \lim_{x\to 2} \frac{x^2 + 3x - 10}{x^2 - 4}$ . This is an indeterminate form $0/0$ . We factor the numerator and denominator: $L = \lim_{x\to 2} \frac{(x-2)(x+5)}{(x-2)(x+2)}$ . We can cancel the $(x-2)$ term for $x \neq 2$ : $L = \lim_{x\to 2} \frac{x+5}{x+2}$ . Now, we can substitute $x=2$ : $L = \frac{2+5}{2+2} = \frac{7}{4}$ .
Evaluate the limit: $\lim_{x \to 2} \frac{x^3 - 2x^2 - 3x + 6}{x^2 - 4}$
Answer: 1/4 The limit is of the indeterminate form $\frac{0}{0}$ . We need to factor the numerator and denominator. The numerator can be factored by grouping: $x^3 - 2x^2 - 3x + 6 = x^2(x-2) - 3(x-2) = (x^2-3)(x-2)$ The denominator is a difference of squares: $x^2 - 4 = (x-2)(x+2)$ . The limit becomes: $\lim_{x \to 2} \frac{(x^2-3)(x-2)}{(x-2)(x+2)}$ Cancel the $(x-2)$ term: $\lim_{x \to 2} \frac{x^2-3}{x+2}$ Substitute $x=2$ : $\frac{2^2-3}{2+2} = \frac{4-3}{4} = \frac{1}{4}$ Distractor B results from a sign error when factoring by grouping, leading to $(x^2+3)(x-2)$ .
Evaluate the limit: $\lim_{x \to -1} \frac{x^{2/3} + x^{1/3}}{x+1}$
Answer: -1/3 The limit is of the indeterminate form $\frac{0}{0}$ . Use the substitution $u = x^{1/3}$ . As $x \to -1$ , $u \to (-1)^{1/3} = -1$ . The expression becomes: $\lim_{u \to -1} \frac{u^2 + u}{u^3+1}$ Factor the numerator and the denominator: $\lim_{u \to -1} \frac{u(u+1)}{(u+1)(u^2-u+1)}$ Cancel the $(u+1)$ term: $\lim_{u \to -1} \frac{u}{u^2-u+1}$ Substitute $u=-1$ : $\frac{-1}{(-1)^2 - (-1) + 1} = \frac{-1}{1+1+1} = -\frac{1}{3}$ Distractor A results from a sign error in the sum of cubes factorization formula in the denominator, using $u^2+u+1$ . Distractor B is a sign error.
Evaluate the limit: $\lim_{x\to \pi/4} \frac{1 - \tan(x)}{\cos(x) - \sin(x)}$
Answer: $\sqrt{2}$ When you encounter a limit that produces an indeterminate form like $\frac{0}{0}$ , you need to use algebraic manipulation or L'Hôpital's rule. Let's first check what happens when we substitute $x = \pi/4$ : both $1 - \tan(\pi/4) = 1 - 1 = 0$ and $\cos(\pi/4) - \sin(\pi/4) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0$ , confirming we have the indeterminate form $\frac{0}{0}$ . The key insight is to factor the denominator cleverly. Notice that $\cos(x) - \sin(x) = -(\sin(x) - \cos(x))$ . Now, multiply both numerator and denominator by the conjugate $(\cos(x) + \sin(x))$ : $\frac{1 - \tan(x)}{\cos(x) - \sin(x)} \cdot \frac{\cos(x) + \sin(x)}{\cos(x) + \sin(x)} = \frac{(1 - \tan(x))(\cos(x) + \sin(x))}{\cos^2(x) - \sin^2(x)}$ Since $\tan(x) = \frac{\sin(x)}{\cos(x)}$ , the numerator becomes $(\cos(x) + \sin(x)) - \sin(x) - \sin^2(x)/\cos(x) = (\cos(x) + \sin(x)) - \sin(x)(1 + \sin(x)/\cos(x)) = \cos(x) - \sin(x)$ Wait, let me use a simpler approach. Factor out $-1$ from the denominator: $\frac{1 - \tan(x)}{-(\sin(x) - \cos(x))} = \frac{-(tan(x) - 1)}{\sin(x) - \cos(x)}$ Using L'Hôpital's rule: $\lim_{x\to \pi/4} \frac{-\sec^2(x)}{-\sin(x) + \cos(x)} = \frac{-2}{-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}} = \sqrt{2}$ This confirms answer D is correct. Answer A (-1) ignores the trigonometric values at $\pi/4$ . Answer B (-√2) has the wrong sign. Answer C (√2/2) likely comes from confusing the trigonometric values themselves with the limit result. Study tip: When facing $\frac{0}{0}$ limits with trigonometric functions, try algebraic manipulation first, then L'Hôpital's rule as backup.
Evaluate the limit: $\lim_{x \to -\infty} \frac{|x^2-5|}{1-2x^2}$
Answer: -1/2 As $x \to -\infty$ , $x^2$ becomes a very large positive number. Thus, $x^2-5$ is positive. Therefore, $|x^2-5| = x^2-5$ for large negative $x$ . The limit becomes: $\lim_{x \to -\infty} \frac{x^2-5}{1-2x^2}$ To evaluate this limit at infinity, we can take the ratio of the coefficients of the highest power terms ( $x^2$ ) in the numerator and denominator. $\frac{1}{-2} = -\frac{1}{2}$ Alternatively, divide every term by $x^2$ : $\lim_{x \to -\infty} \frac{1 - \frac{5}{x^2}}{\frac{1}{x^2}-2} = \frac{1-0}{0-2} = -\frac{1}{2}$ Distractor A is a common error from assuming that the absolute value makes the entire result positive, or from incorrectly evaluating $|x^2-5| = -(x^2-5)$ .
Evaluate the limit: $\lim_{x \to -\infty} (x + \sqrt{x^2+6x})$
Answer: -3 This limit is of the indeterminate form $-\infty + \infty$ . Multiply by the conjugate, $x - \sqrt{x^2+6x}$ , over itself. $\lim_{x \to -\infty} \frac{(x + \sqrt{x^2+6x})(x - \sqrt{x^2+6x})}{x - \sqrt{x^2+6x}}$ $= \lim_{x \to -\infty} \frac{x^2 - (x^2+6x)}{x - \sqrt{x^2+6x}} = \lim_{x \to -\infty} \frac{-6x}{x - \sqrt{x^2+6x}}$ Divide the numerator and denominator by $x$ . Since $x \to -\infty$ , $\sqrt{x^2} = |x| = -x$ . $= \lim_{x \to -\infty} \frac{\frac{-6x}{x}}{\frac{x}{x} - \frac{\sqrt{x^2(1+6/x)}}{x}} = \lim_{x \to -\infty} \frac{-6}{1 - \frac{|x|\sqrt{1+6/x}}{x}}$ $= \lim_{x \to -\infty} \frac{-6}{1 - \frac{-x\sqrt{1+6/x}}{x}} = \lim_{x \to -\infty} \frac{-6}{1 + \sqrt{1+6/x}}$ As $x \to -\infty$ , $\frac{6}{x} \to 0$ : $\frac{-6}{1+\sqrt{1+0}} = \frac{-6}{1+1} = -3$ Distractor B is a sign error. Distractor D is chosen if one forgets that $\sqrt{x^2}=-x$ for negative $x$ , leading to a zero in the denominator.
Evaluate the limit: $\lim_{x\to 8} \frac{\sqrt[3]{x} - 2}{x - 8}$
Answer: $1/12$ This limit is in the indeterminate form $0/0$ . We can resolve this by using the algebraic identity for the difference of cubes, $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ . Let $a = \sqrt[3]{x}$ and $b = 2$ . We multiply the numerator and denominator by the 'conjugate' term $((\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4)$ : $\lim_{x\to 8} \frac{\sqrt[3]{x} - 2}{x - 8} \cdot \frac{(\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4}{(\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4} = \lim_{x\to 8} \frac{x - 8}{(x - 8)((\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4)}$ After canceling the $(x-8)$ terms, we evaluate the limit: $\lim_{x\to 8} \frac{1}{(\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4} = \frac{1}{(\sqrt[3]{8})^2 + 2\sqrt[3]{8} + 4} = \frac{1}{2^2 + 2(2) + 4} = \frac{1}{4+4+4} = \frac{1}{12}$ Distractor B, $1/8$ , might result from an error in evaluating the denominator as $2+2+4=8$ . Distractor C, $1/4$ , might result from only considering one of the three terms in the denominator. Distractor D, $0$ , is the result of incorrect direct substitution into the numerator.
Evaluate the limit: $\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x^3+8}$
Answer: -1/48 This limit is of the indeterminate form $\frac{0}{0}$ . First, simplify the complex fraction in the numerator. $\frac{1}{x} + \frac{1}{2} = \frac{2+x}{2x}$ The expression becomes: $\lim_{x \to -2} \frac{\frac{x+2}{2x}}{x^3+8}$ Factor the denominator as a sum of cubes: $x^3+8 = (x+2)(x^2-2x+4)$ . $\lim_{x \to -2} \frac{x+2}{2x(x+2)(x^2-2x+4)}$ Cancel the $(x+2)$ term: $\lim_{x \to -2} \frac{1}{2x(x^2-2x+4)}$ Substitute $x=-2$ : $\frac{1}{2(-2)((-2)^2-2(-2)+4)} = \frac{1}{-4(4+4+4)} = \frac{1}{-4(12)} = -\frac{1}{48}$ Distractor A results from forgetting the $2x$ term from the initial simplification. Distractor B results from a sign error in the sum of cubes formula.
What is the value of the limit $\lim_{x\to 4} \frac{\sqrt{x} - 2}{x^2 - 5x + 4}$ ?
Answer: $1/12$ When you encounter a limit that produces the indeterminate form $\frac{0}{0}$ , you need to algebraically manipulate the expression before evaluating. Let's check: as $x \to 4$ , the numerator becomes $\sqrt{4} - 2 = 0$ and the denominator becomes $16 - 20 + 4 = 0$ , confirming we have $\frac{0}{0}$ . To resolve this, we'll rationalize the numerator by multiplying by the conjugate $\frac{\sqrt{x} + 2}{\sqrt{x} + 2}$ : $\frac{\sqrt{x} - 2}{x^2 - 5x + 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{(\sqrt{x})^2 - 4}{(x^2 - 5x + 4)(\sqrt{x} + 2)} = \frac{x - 4}{(x^2 - 5x + 4)(\sqrt{x} + 2)}$ Next, factor the denominator's quadratic: $x^2 - 5x + 4 = (x - 4)(x - 1)$ . This gives us: $\frac{x - 4}{(x - 4)(x - 1)(\sqrt{x} + 2)} = \frac{1}{(x - 1)(\sqrt{x} + 2)}$ Now we can evaluate the limit: $\lim_{x\to 4} \frac{1}{(x - 1)(\sqrt{x} + 2)} = \frac{1}{(4 - 1)(\sqrt{4} + 2)} = \frac{1}{3 \cdot 4} = \frac{1}{12}$ Answer (A) $\frac{1}{8}$ results from miscalculating the denominator as $4 \cdot 2 = 8$ . Answer (B) $\frac{1}{3}$ comes from forgetting the $(\sqrt{x} + 2)$ factor. Answer (C) $\frac{1}{4}$ occurs when you forget the $(x - 1)$ factor. Study tip: When facing $\frac{0}{0}$ limits involving square roots, always try rationalizing the numerator first. Factor completely before canceling to avoid missing terms.
Evaluate the limit: $\lim_{x \to 1} \frac{x^4 - 1}{x^3 - 1}$
Answer: 4/3 The limit is of the indeterminate form $\frac{0}{0}$ . We factor the numerator and the denominator. Numerator (difference of squares twice): $x^4 - 1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)$ . Denominator (difference of cubes): $x^3 - 1 = (x-1)(x^2+x+1)$ . The limit becomes: $\lim_{x \to 1} \frac{(x-1)(x+1)(x^2+1)}{(x-1)(x^2+x+1)}$ Cancel the $(x-1)$ term: $\lim_{x \to 1} \frac{(x+1)(x^2+1)}{x^2+x+1}$ Substitute $x=1$ : $\frac{(1+1)(1^2+1)}{1^2+1+1} = \frac{2 \cdot 2}{3} = \frac{4}{3}$ Distractor D might result from an incorrect factorization. Distractor C is the reciprocal of the answer.
Let $a$ be a non-zero constant. Evaluate $\lim_{x\to a} \frac{x^2 - (a+1)x + a}{x^3 - a^3}$
Answer: $\frac{a-1}{3a^2}$ When you encounter a limit that produces the indeterminate form $\frac{0}{0}$ , you need to factor and simplify before evaluating. Let's substitute $x = a$ into both numerator and denominator to confirm we get $\frac{0}{0}$ . For the numerator $x^2 - (a+1)x + a$ , we need to factor this quadratic. Since it equals zero when $x = a$ , we know $(x-a)$ is a factor. Using polynomial division or factoring: $x^2 - (a+1)x + a = (x-a)(x-1)$ . For the denominator $x^3 - a^3$ , we use the difference of cubes formula: $x^3 - a^3 = (x-a)(x^2 + ax + a^2)$ . Now our limit becomes: $\lim_{x\to a} \frac{(x-a)(x-1)}{(x-a)(x^2 + ax + a^2)} = \lim_{x\to a} \frac{x-1}{x^2 + ax + a^2}$ Since we've removed the common factor $(x-a)$ , we can now substitute $x = a$ : $\frac{a-1}{a^2 + a \cdot a + a^2} = \frac{a-1}{3a^2}$ This confirms answer C is correct. Looking at the wrong answers: A) $\frac{1}{3a}$ ignores the numerator's contribution entirely. B) $\frac{a+1}{3a^2}$ incorrectly uses $(a+1)$ instead of $(a-1)$ in the numerator—a sign error from the factoring. D) $\frac{1}{a^2}$ has the wrong denominator, suggesting incorrect handling of the cubic term. Strategy tip: Always factor completely when you get $\frac{0}{0}$ . Check your factoring by expanding back, and remember that $x^3 - a^3 = (x-a)(x^2 + ax + a^2)$ .
What is the value of the limit $\lim_{h\to 0} \frac{\frac{1}{(h+3)^2} - \frac{1}{9}}{h}$ ?
Answer: $-2/27$ This limit is of the form $0/0$ . To evaluate it, we simplify the complex fraction in the numerator by finding a common denominator: $\frac{1}{(h+3)^2} - \frac{1}{9} = \frac{9 - (h+3)^2}{9(h+3)^2} = \frac{9 - (h^2+6h+9)}{9(h+3)^2} = \frac{-h^2-6h}{9(h+3)^2} = \frac{-h(h+6)}{9(h+3)^2}$ Now, substitute this back into the original limit expression: $\lim_{h\to 0} \frac{\frac{-h(h+6)}{9(h+3)^2}}{h} = \lim_{h\to 0} \frac{-h(h+6)}{9h(h+3)^2}$ We can cancel the $h$ term: $\lim_{h\to 0} \frac{-(h+6)}{9(h+3)^2}$ Now, substitute $h=0$ : $\frac{-(0+6)}{9(0+3)^2} = \frac{-6}{9(9)} = \frac{-6}{81} = -\frac{2}{27}$ Distractor B reflects a sign error. Distractors C and D result from errors in expanding $(h+3)^2$ or simplifying the numerator.
Evaluate the limit: $\lim_{x \to 0} \frac{\frac{1}{(x+2)^2} - \frac{1}{4}}{x}$
Answer: -1/4 The limit is of the indeterminate form $\frac{0}{0}$ . First, simplify the complex fraction in the numerator by finding a common denominator. $\frac{\frac{4 - (x+2)^2}{4(x+2)^2}}{x} = \frac{4 - (x^2+4x+4)}{4x(x+2)^2}$ $= \frac{-x^2-4x}{4x(x+2)^2} = \frac{-x(x+4)}{4x(x+2)^2}$ The limit becomes: $\lim_{x \to 0} \frac{-x(x+4)}{4x(x+2)^2}$ Cancel the $x$ term: $\lim_{x \to 0} \frac{-(x+4)}{4(x+2)^2}$ Substitute $x=0$ : $\frac{-(0+4)}{4(0+2)^2} = \frac{-4}{4(4)} = -\frac{1}{4}$ Distractor D results from a sign error when expanding $-(x+2)^2$ . Distractor C results from forgetting the denominator $4(x+2)^2$ in the simplification.
Evaluate the limit: $\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x-8}$
Answer: 1/12 The limit is of the indeterminate form $\frac{0}{0}$ . We can use the formula for the difference of cubes, $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ . Let $a = \sqrt[3]{x}$ and $b=2$ . Then the denominator $x-8 = (\sqrt[3]{x})^3 - 2^3$ . $x-8 = (\sqrt[3]{x} - 2)((\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4)$ So the limit becomes: $\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{(\sqrt[3]{x} - 2)((\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4)}$ Cancel the $(\sqrt[3]{x} - 2)$ term: $\lim_{x \to 8} \frac{1}{(\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4}$ Substitute $x=8$ : $\frac{1}{(\sqrt[3]{8})^2 + 2\sqrt[3]{8} + 4} = \frac{1}{2^2 + 2(2) + 4} = \frac{1}{4+4+4} = \frac{1}{12}$ Distractor B results from a sign error in the cube factorization formula. Distractor C is the reciprocal of the correct answer.
Evaluate the limit: $\lim_{x \to 1} \frac{(\sqrt{x}-1)^2}{x-1}$
Answer: 0 The limit is of the indeterminate form $\frac{0}{0}$ . The key is to factor the denominator as a difference of squares: $x-1 = (\sqrt{x}-1)(\sqrt{x}+1)$ . The expression becomes: $\lim_{x \to 1} \frac{(\sqrt{x}-1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}$ Cancel one factor of $(\sqrt{x}-1)$ from the numerator and denominator: $\lim_{x \to 1} \frac{\sqrt{x}-1}{\sqrt{x}+1}$ Now, substitute $x=1$ : $\frac{\sqrt{1}-1}{\sqrt{1}+1} = \frac{1-1}{1+1} = \frac{0}{2} = 0$ Distractor C results from incorrectly canceling the numerator, leaving 1 instead of $(\sqrt{x}-1)$ .
Let $f$ be the function defined by $f(x) = \begin{cases} \frac{x-2}{\sqrt{x-1}-1} & \text{if } x > 2 \\ ax + 1 & \text{if } x \le 2 \end{cases}$ . For what value of the constant $a$ is the function $f$ continuous at $x = 2$ ?
Answer: $1/2$ When you encounter a piecewise function and need to find a constant that makes it continuous at a boundary point, you're testing whether the left-hand and right-hand limits equal the function value at that point. For continuity at $x = 2$ , we need $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$ . First, find the left-hand limit. Since $x \leq 2$ uses $f(x) = ax + 1$ , we have $\lim_{x \to 2^-} f(x) = a(2) + 1 = 2a + 1$ . Also, $f(2) = 2a + 1$ . Next, find the right-hand limit using $f(x) = \frac{x-2}{\sqrt{x-1}-1}$ for $x > 2$ . This gives the indeterminate form $\frac{0}{0}$ as $x \to 2^+$ , so we need to simplify. Multiply by the conjugate: $\frac{x-2}{\sqrt{x-1}-1} \cdot \frac{\sqrt{x-1}+1}{\sqrt{x-1}+1} = \frac{(x-2)(\sqrt{x-1}+1)}{(x-1)-1} = \frac{(x-2)(\sqrt{x-1}+1)}{x-2} = \sqrt{x-1}+1$ Therefore, $\lim_{x \to 2^+} f(x) = \sqrt{2-1}+1 = 1+1 = 2$ . Setting the limits equal: $2a + 1 = 2$ , so $a = \frac{1}{2}$ . Looking at the wrong answers: A) $a = 1$ gives $2a + 1 = 3 \neq 2$ . B) $a = 2$ gives $2a + 1 = 5 \neq 2$ . D) $a = -\frac{1}{2}$ gives $2a + 1 = 0 \neq 2$ . Study tip: For piecewise continuity problems, always rationalize expressions with square roots when you encounter $\frac{0}{0}$ forms—the conjugate method is your key tool.
For a function defined by $f(x) = \sqrt{x}$ , where $x>0$ , which of the following expressions represents $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ ?
Answer: \frac{1}{2\sqrt{x}} The question asks to evaluate the limit $\lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$ . This is an indeterminate form $\frac{0}{0}$ . We multiply the numerator and denominator by the conjugate, $\sqrt{x+h} + \sqrt{x}$ . $\lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}$ $= \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$ $= \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$ Cancel the $h$ terms: $= \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$ Substitute $h=0$ : $= \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$ Distractor A results from forgetting the conjugate in the denominator. Distractor B is the reciprocal of the correct answer, with the radical incorrectly placed.
Evaluate the limit: $\lim_{x \to 1} \frac{x^{1/2} - 1}{x^{1/3} - 1}$
Answer: 3/2 The limit is of the indeterminate form $\frac{0}{0}$ . To simplify, use a substitution. Let $u = x^{1/6}$ . Then $x^{1/2} = (x^{1/6})^3 = u^3$ and $x^{1/3} = (x^{1/6})^2 = u^2$ . As $x \to 1$ , $u \to 1^{1/6} = 1$ . The limit in terms of $u$ is: $\lim_{u \to 1} \frac{u^3 - 1}{u^2 - 1}$ Factor the numerator and denominator: $\lim_{u \to 1} \frac{(u-1)(u^2+u+1)}{(u-1)(u+1)}$ Cancel the $(u-1)$ term: $\lim_{u \to 1} \frac{u^2+u+1}{u+1}$ Substitute $u=1$ : $\frac{1^2+1+1}{1+1} = \frac{3}{2}$ Distractor B inverts the correct answer, a common mistake related to the exponents. Distractor A could result from a simplification error.
Evaluate the limit: $\lim_{x \to 3^-} \frac{x^2 - 9}{|2x-6|}$
Answer: -3 This is a one-sided limit involving an absolute value. As $x \to 3^-$ , $x$ is slightly less than 3. Therefore, the term inside the absolute value, $2x-6$ , is slightly negative. By definition of absolute value, $|2x-6| = -(2x-6)$ for $x<3$ . The limit becomes: $\lim_{x \to 3^-} \frac{x^2 - 9}{-(2x-6)}$ Factor the numerator and the denominator: $\lim_{x \to 3^-} \frac{(x-3)(x+3)}{-2(x-3)}$ Cancel the $(x-3)$ term: $\lim_{x \to 3^-} \frac{x+3}{-2}$ Substitute $x = 3$ : $\frac{3+3}{-2} = \frac{6}{-2} = -3$ Distractor B is the value of the limit as $x \to 3^+$ , a common mistake from ignoring the one-sided direction of the limit.
Evaluate the limit: $\lim_{x \to 0} \frac{1}{x} \left( \frac{1}{2+x} - \frac{1}{2} \right)$
Answer: -1/4 The expression is of the indeterminate form $\infty \cdot 0$ . First, combine the fractions inside the parentheses. $\frac{1}{2+x} - \frac{1}{2} = \frac{2 - (2+x)}{2(2+x)} = \frac{-x}{2(2+x)}$ Now substitute this back into the limit expression: $\lim_{x \to 0} \frac{1}{x} \left( \frac{-x}{2(2+x)} \right)$ Cancel the $x$ terms: $\lim_{x \to 0} \frac{-1}{2(2+x)}$ Substitute $x=0$ : $\frac{-1}{2(2+0)} = -\frac{1}{4}$ Distractor A results from forgetting the denominator $2(2+x)$ after simplification. Distractor B is a sign error.

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