This limit creates 0/0, requiring factorization of the difference of squares. The numerator x² - 1 factors as (x - 1)(x + 1). The expression becomes (x - 1)(x + 1)/(x + 1), which simplifies to (x - 1) after canceling the common factor (x + 1). Substituting x = -1 gives -1 - 1 = -2. A common error is not recognizing the difference of squares pattern or incorrectly canceling factors. The transferable method is to identify special polynomial forms, factor completely, cancel common terms that create indeterminate forms, and evaluate by substitution.

This limit produces 0/0, requiring factorization of the numerator by extracting the common factor. The expression $x^2 + 2x$ factors as $x(x + 2)$. The limit becomes $\frac{x(x + 2)}{x + 2}$, which simplifies to $x$ after canceling $(x + 2)$. Substituting $x = -2$ gives $-2$. Students often don't recognize the factorization or make sign errors during substitution. The essential technique is to factor polynomials by identifying common factors, cancel terms that create indeterminate forms, and evaluate by direct substitution.

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.

This limit has the indeterminate form 0/0, requiring us to factor the numerator. We can factor x³ - 1 using the difference of cubes formula: a³ - b³ = (a - b)(a² + ab + b²). Here, x³ - 1 = (x - 1)(x² + x + 1). The expression becomes (x - 1)(x² + x + 1)/(x - 1), and the common factor (x - 1) cancels. We're left with x² + x + 1, and substituting x = 1 gives 1² + 1 + 1 = 3. A common mistake is forgetting the difference of cubes formula or incorrectly applying it. The transferable strategy is to memorize factoring formulas for special forms like difference of cubes: a³ - b³ = (a - b)(a² + ab + b²).

This limit gives 0/0 when substituting x = 1, requiring algebraic manipulation. The numerator 1 - x² can be rewritten as -(x² - 1) = -(x - 1)(x + 1). The denominator 1 - x can be rewritten as -(x - 1). Our limit becomes lim[x→1] [-(x - 1)(x + 1)]/[-(x - 1)] = lim[x→1] (x + 1) after canceling -(x - 1) from both numerator and denominator. Substituting x = 1 gives 1 + 1 = 2. A common error is not recognizing that 1 - x² = -(x² - 1) or forgetting to handle the negative signs properly when factoring. When you see expressions like 1 - x² or a - b, consider rewriting them as -(x² - 1) or -(b - a) to reveal familiar factoring patterns.

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.

Direct substitution gives 0/0, so we need to factor both numerator and denominator. The numerator x² - 4 is a difference of squares: (x - 2)(x + 2). The denominator x² - 2x factors as x(x - 2). Our limit becomes lim[x→2] [(x - 2)(x + 2)]/[x(x - 2)], and after canceling (x - 2), we get lim[x→2] (x + 2)/x. Substituting x = 2 gives (2 + 2)/2 = 4/2 = 2. Students sometimes factor only the numerator and forget about the denominator, or make errors when factoring out common terms. Always factor completely before canceling to avoid missing simplifications.

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.

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.

This limit produces 0/0, requiring factorization of the numerator. The expression x² + 3x factors as x(x + 3). The limit becomes x(x + 3)/(x + 3), which simplifies to x after canceling the common factor (x + 3). Substituting x = -3 gives -3. A common mistake is not factoring the numerator completely or incorrectly canceling terms. The essential strategy is to factor polynomials by identifying common factors, cancel terms that cause the indeterminate form, and then evaluate by direct substitution.