Determining Limits Using Algebraic Manipulation - AP Calculus AB

A formula includes $\frac{\sqrt{x+1}-2}{x-3}$ ; compute $\displaystyle\lim_{x\to 3}\frac{\sqrt{x+1}-2}{x-3}$ .
Answer: $\frac{1}{4}$ This limit gives 0/0 when x = 3, involving a radical expression. We rationalize by multiplying by the conjugate (√(x+1) + 2)/(√(x+1) + 2). This gives [(√(x+1) - 2)(√(x+1) + 2)]/[(x-3)(√(x+1) + 2)] = (x+1-4)/[(x-3)(√(x+1) + 2)] = (x-3)/[(x-3)(√(x+1) + 2)]. After canceling (x-3), we have 1/(√(x+1) + 2), and substituting x = 3 gives 1/(√4 + 2) = 1/(2 + 2) = 1/4. Students often make arithmetic errors when expanding (√(x+1))² or forget to add 2 in the denominator after substitution. The strategy: multiply by the conjugate, simplify the numerator completely, then cancel and evaluate.
Compute $\lim\limits_{x\to 5}\dfrac{x^2-25}{x-5}$ , an indeterminate form that simplifies by factoring.
Answer: $10$ This limit involves the indeterminate form 0/0, requiring factorization of the difference of squares. The numerator x² - 25 factors as (x - 5)(x + 5). The expression becomes (x - 5)(x + 5)/(x - 5), which simplifies to (x + 5) after canceling the common factor (x - 5). Substituting x = 5 gives 5 + 5 = 10. A frequent error is not recognizing the difference of squares pattern or incorrectly factoring. The systematic strategy is to identify special polynomial forms like difference of squares, factor completely, cancel common terms, and then evaluate by substitution.
Determine $\lim\limits_{x\to 3}\dfrac{x^2-3x}{x-3}$ by simplifying first.
Answer: $3$ This limit produces 0/0, requiring factorization of the numerator by extracting the common factor. The expression x² - 3x factors as x(x - 3). The limit becomes x(x - 3)/(x - 3), which simplifies to x after canceling (x - 3). Substituting x = 3 gives 3. A frequent error is not factoring the numerator completely or making cancellation mistakes. The systematic approach is to factor polynomials by identifying common factors, cancel terms that create the indeterminate form, and evaluate by substitution.
A model uses $\dfrac{\sqrt{x+9}-3}{x}$ ; determine $\lim\limits_{x\to 0}\dfrac{\sqrt{x+9}-3}{x}$ .
Answer: $\dfrac{1}{6}$ This limit involves a radical expression that creates 0/0, requiring rationalization of the numerator. Multiply both numerator and denominator by the conjugate ( $\sqrt{x+9} + 3$ ). The numerator becomes ( $\sqrt{x+9} - 3$ )( $\sqrt{x+9} + 3$ ) = (x+9) - 9 = x. The expression simplifies to $\frac{x}{x(\sqrt{x+9} + 3)}$ = $\frac{1}{\sqrt{x+9} + 3}$ after canceling x. Substituting x = 0 gives $\frac{1}{\sqrt{9} + 3}$ = $\frac{1}{3 + 3}$ = $\frac{1}{6}$ . Students often forget to rationalize or make algebraic errors during the process. The key strategy for radical limits is to rationalize using the conjugate to eliminate the radical from the numerator.
Compute $\lim\limits_{x\to 2}\dfrac{x^2+x-6}{x-2}$ after simplifying the $\frac{0}{0}$ form.
Answer: $5$ This limit creates 0/0, requiring factorization of the quadratic numerator. The expression x² + x - 6 factors as (x + 3)(x - 2) by finding two numbers that multiply to -6 and add to 1. The limit becomes (x + 3)(x - 2)/(x - 2), which simplifies to (x + 3) after canceling the common factor (x - 2). Substituting x = 2 gives 2 + 3 = 5. A frequent error is incorrectly factoring the quadratic or making substitution mistakes. The essential strategy is to factor quadratic expressions completely, identify and cancel common factors, then evaluate by direct substitution.
A function is defined by $f(x)=\frac{\sqrt{x+9}-3}{x}$ for $x\ne0$ ; find $\lim_{x\to0}f(x)$ .
Answer: $\frac{1}{6}$ This limit has the indeterminate form 0/0, requiring rationalization since we have a square root. We multiply both numerator and denominator by the conjugate (√(x + 9) + 3). The numerator becomes (√(x + 9) - 3)(√(x + 9) + 3) = (x + 9) - 9 = x using the difference of squares. The denominator becomes x(√(x + 9) + 3), so our expression simplifies to x/[x(√(x + 9) + 3)] = 1/(√(x + 9) + 3). Substituting x = 0 gives 1/(√9 + 3) = 1/(3 + 3) = 1/6. A common error is forgetting to multiply the denominator by the conjugate or making arithmetic mistakes. The key strategy for limits with square roots is to rationalize using conjugates.
Evaluate $\lim_{x\to 9}\frac{\sqrt{x}-3}{x-9}$ by rationalizing to eliminate the $0/0$ form.
Answer: $\frac{1}{6}$ This limit has 0/0 form, and with a square root in the numerator, we should rationalize. Multiply numerator and denominator by the conjugate (√x + 3). The numerator becomes (√x - 3)(√x + 3) = x - 9, while the denominator becomes (x - 9)(√x + 3). This gives us (x - 9)/[(x - 9)(√x + 3)], and the common factor (x - 9) cancels. We're left with 1/(√x + 3), and substituting x = 9 gives 1/(√9 + 3) = 1/(3 + 3) = 1/6. A common error is rationalizing incorrectly or making arithmetic mistakes with the conjugate. The key strategy for limits with radicals is to multiply by the conjugate to eliminate the radical from the numerator.
Compute $\displaystyle\lim_{x\to 1}\frac{x^3-1}{x-1}$ by algebraic manipulation to remove the $\frac{0}{0}$ form.
Answer: $3$ Direct substitution yields 0/0, so we need to factor x³ - 1. This is a difference of cubes with the formula a³ - b³ = (a - b)(a² + ab + b²). Here, x³ - 1 = (x - 1)(x² + x + 1), giving us lim[x→1] [(x - 1)(x² + x + 1)]/(x - 1). Canceling the common factor (x - 1) leaves lim[x→1] (x² + x + 1) = 1² + 1 + 1 = 3. Students often forget the difference of cubes formula or make errors in the middle term (it's +ab, not -ab). Remember: for x³ - a³, factor as (x - a)(x² + ax + a²), then cancel and evaluate.
A particle's average velocity is modeled by $\lim_{t\to 3}\frac{t^2-9}{t-3}$ . What is the limit?
Answer: $6$ This limit has the indeterminate form 0/0 when we substitute t = 3, so we need algebraic manipulation. The numerator t² - 9 is a difference of squares that factors as (t - 3)(t + 3). After factoring, we get lim[t→3] [(t - 3)(t + 3)]/(t - 3), and the (t - 3) terms cancel out, leaving lim[t→3] (t + 3). Now we can directly substitute t = 3 to get 3 + 3 = 6. A common error is trying to apply L'Hôpital's rule immediately without first attempting algebraic simplification. The key strategy is to recognize special factoring patterns like difference of squares (a² - b² = (a - b)(a + b)) before considering more advanced techniques.
A temperature adjustment uses $\lim_{h\to 0}\frac{(5+h)^2-25}{h}$ . What is the value of the limit?
Answer: $10$ This limit represents a difference quotient form. Expanding (5 + h)² gives 25 + 10h + h², so the numerator becomes 25 + 10h + h² - 25 = 10h + h². We can factor out h from the numerator to get h(10 + h), giving us lim[h→0] [h(10 + h)]/h. After canceling the h terms, we have lim[h→0] (10 + h), which equals 10 when h = 0. Students often make errors when expanding the squared binomial or forget to distribute the negative sign. This form appears frequently when finding derivatives using the definition, so recognizing it saves time.
A particle's average velocity is $\dfrac{(x+1)^2-4}{x-1}$ ; find $\lim\limits_{x\to 1}\dfrac{(x+1)^2-4}{x-1}$ .
Answer: $4$ This limit presents an indeterminate form 0/0 that requires algebraic simplification. First, expand the numerator: (x+1)² - 4 = x² + 2x + 1 - 4 = x² + 2x - 3. Factor this quadratic as (x + 3)(x - 1). The expression becomes (x + 3)(x - 1)/(x - 1), which simplifies to (x + 3) after canceling the common factor. Substituting x = 1 yields 1 + 3 = 4. A frequent mistake is expanding incorrectly or not recognizing the factored form. The transferable approach is to expand, factor, and cancel common terms that cause the indeterminate form.
Compute $\lim\limits_{x\to 0}\dfrac{(3+x)^2-9}{x}$ by simplifying the expression.
Answer: $6$ This limit creates 0/0, requiring expansion and simplification of the squared binomial. Expand (3+x)²: (3+x)² = 9 + 6x + x². The numerator becomes 9 + 6x + x² - 9 = 6x + x² = x(6 + x). The expression becomes x(6 + x)/x, which simplifies to (6 + x) after canceling x. Substituting x = 0 gives 6 + 0 = 6. Students often make errors expanding binomial squares or not factoring out x correctly. The strategy is to expand carefully, factor common terms, cancel appropriately, and substitute the limit value.
An average rate is given by $\lim_{x\to 3}\frac{2x^2-18}{x-3}$ . Find the limit after factoring.
Answer: $12$ Direct substitution yields 0/0, so we need to factor the numerator. First, factor out 2 from 2x² - 18 to get 2(x² - 9). Then x² - 9 is a difference of squares that factors as (x - 3)(x + 3). Our limit becomes lim[x→3] [2(x - 3)(x + 3)]/(x - 3), and after canceling (x - 3), we have lim[x→3] 2(x + 3). Substituting x = 3 gives 2(3 + 3) = 2(6) = 12. A common mistake is forgetting to factor out the common coefficient before applying difference of squares, leading to an answer of 6 instead of 12. Always look for common factors first before applying other factoring techniques.
For a rate near $x=0$ , find $\lim\limits_{x\to 0}\dfrac{\sqrt{1+x}-1}{x}$ .
Answer: $\dfrac{1}{2}$ This limit involves a radical expression creating 0/0, requiring rationalization of the numerator. Multiply both numerator and denominator by the conjugate $(\sqrt{1+x} + 1)$ . The numerator becomes $(\sqrt{1+x} - 1)(\sqrt{1+x} + 1) = (1+x) - 1 = x$ . The expression simplifies to $\frac{x}{x(\sqrt{1+x} + 1)} = \frac{1}{\sqrt{1+x} + 1}$ after canceling x. Substituting x = 0 gives $\frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$ . Students often forget to use the conjugate or make errors during rationalization. The essential technique for radical limits is to multiply by the conjugate to eliminate radicals and create factorable forms.
Compute $\lim\limits_{x\to -1}\dfrac{x^2+2x+1}{x+1}$ , which is indeterminate before simplifying.
Answer: $0$ This limit involves 0/0, requiring recognition of a perfect square trinomial in the numerator. The expression x² + 2x + 1 factors as (x + 1)² or (x + 1)(x + 1). The limit becomes (x + 1)²/(x + 1), which simplifies to (x + 1) after canceling one factor of (x + 1). Substituting x = -1 gives -1 + 1 = 0. A common mistake is not recognizing the perfect square trinomial or incorrectly canceling both factors. The strategy is to identify perfect square patterns, factor appropriately, cancel one instance of common factors, and substitute carefully.
Evaluate $\lim\limits_{x\to 0}\dfrac{\sqrt{4+x}-2}{x}$ by rationalizing the numerator.
Answer: $\dfrac{1}{4}$ This limit involves a radical creating 0/0, requiring rationalization using the conjugate. Multiply both numerator and denominator by (√(4+x) + 2). The numerator becomes (√(4+x) - 2)(√(4+x) + 2) = (4+x) - 4 = x. The expression simplifies to x/(x(√(4+x) + 2)) = 1/(√(4+x) + 2) after canceling x. Substituting x = 0 gives 1/(√4 + 2) = 1/(2 + 2) = 1/4. A common error is not recognizing the need for rationalization or making algebraic mistakes. The strategy for radical expressions is to rationalize using conjugates, simplify algebraically, and then evaluate by substitution.
Evaluate $\lim\limits_{x\to 1}\dfrac{2x^2-2}{x-1}$ using algebraic simplification.
Answer: $4$ This limit creates 0/0, requiring factorization of the numerator by extracting common factors. The expression 2x² - 2 factors as 2(x² - 1) = 2(x - 1)(x + 1) using the difference of squares. The limit becomes 2(x - 1)(x + 1)/(x - 1), which simplifies to 2(x + 1) after canceling (x - 1). Substituting x = 1 gives 2(1 + 1) = 2(2) = 4. A common error is not factoring completely or making mistakes during the cancellation process. The strategy is to factor systematically, identify common factors, cancel appropriately, and substitute.
A rate is defined by $\frac{\sqrt{x+5}-3}{x-4}$ ; find $\displaystyle\lim_{x\to 4}\frac{\sqrt{x+5}-3}{x-4}$ .
Answer: $\frac{1}{6}$ This limit involves a radical expression creating 0/0 when x = 4. To resolve this, we multiply by the conjugate (√(x+5) + 3) over itself to rationalize. This gives us [(√(x+5) - 3)(√(x+5) + 3)]/[(x-4)(√(x+5) + 3)] = (x+5-9)/[(x-4)(√(x+5) + 3)] = (x-4)/[(x-4)(√(x+5) + 3)]. After canceling (x-4), we get 1/(√(x+5) + 3), and substituting x = 4 yields 1/(√9 + 3) = 1/(3 + 3) = 1/6. A common error is forgetting to multiply both numerator and denominator by the conjugate. The key strategy: when radicals create 0/0, use conjugate multiplication to eliminate the radical difference.
Evaluate $\lim\limits_{x\to 1}\dfrac{x^2-3x+2}{x-1}$ , an indeterminate form removable by factoring.
Answer: $-1$ This limit initially gives 0/0, requiring factorization of the quadratic numerator. The expression x² - 3x + 2 factors as (x - 1)(x - 2) by finding two numbers that multiply to 2 and add to -3. The limit becomes (x - 1)(x - 2)/(x - 1), which simplifies to (x - 2) after canceling the common factor (x - 1). Substituting x = 1 gives 1 - 2 = -1. A common error is incorrectly factoring the quadratic or making substitution mistakes. The transferable technique is to factor quadratics systematically, cancel common factors that create indeterminate forms, and evaluate carefully.
Find $\displaystyle\lim_{x\to 2}\frac{x^2+x-6}{x-2}$ using algebra to resolve the indeterminate form.
Answer: $5$ Substituting x = 2 gives 0/0, so we factor the numerator x² + x - 6. We need two numbers that multiply to -6 and add to 1, which are 3 and -2. Thus x² + x - 6 = (x + 3)(x - 2), and our limit becomes lim[x→2] [(x + 3)(x - 2)]/(x - 2). Canceling the (x - 2) factor leaves lim[x→2] (x + 3) = 2 + 3 = 5. A common error is incorrect factoring—students might try (x + 2)(x - 3) which gives x² - x - 6 instead. Always verify your factoring by expanding, then cancel and substitute to find the limit.

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