This limit has the indeterminate form 0/0 when we substitute x = 3 directly, so we need algebraic manipulation. The numerator x² - 9 is a difference of squares that factors as (x - 3)(x + 3). After factoring, we get lim[x→3] [(x - 3)(x + 3)]/(x - 3), and since x approaches but never equals 3, we can cancel the (x - 3) terms. This leaves us with lim[x→3] (x + 3) = 3 + 3 = 6. A common error is thinking the limit doesn't exist because of the 0/0 form, but factoring reveals the removable discontinuity. The strategy is: when you see 0/0, look for common factors to cancel before evaluating.

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.

This limit creates 0/0, requiring rationalization of the denominator since it contains a radical. Multiply both numerator and denominator by the conjugate (√x + 1). The denominator becomes (√x - 1)(√x + 1) = x - 1. The expression becomes (1 - x)(√x + 1)/(x - 1) = -(x - 1)(√x + 1)/(x - 1) = -(√x + 1) after canceling (x - 1). Substituting x = 1 gives -(√1 + 1) = -(1 + 1) = -2. Students often struggle with rationalizing denominators or handling the negative sign properly. The technique is to rationalize denominators with conjugates and track signs carefully.

This limit creates 0/0, requiring factorization using the difference of cubes formula. The numerator x³ - 1 factors as (x - 1)(x² + x + 1) using the pattern a³ - b³ = (a - b)(a² + ab + b²). The expression becomes (x - 1)(x² + x + 1)/(x - 1), which simplifies to x² + x + 1 after canceling (x - 1). Substituting x = 1 gives 1² + 1 + 1 = 3. Students often don't recognize the difference of cubes pattern or make errors applying the formula. The key technique is to identify cubic patterns and apply the appropriate factorization formulas systematically.

This limit involves a radical expression creating 0/0, requiring rationalization of the numerator. Multiply both numerator and denominator by the conjugate (√(x+3) + 2). The numerator becomes (√(x+3) - 2)(√(x+3) + 2) = (x+3) - 4 = x - 1. The expression becomes (x - 1)/((x - 1)(√(x+3) + 2)), which simplifies to 1/(√(x+3) + 2) after canceling (x - 1). Substituting x = 1 gives 1/(√(1+3) + 2) = 1/(√4 + 2) = 1/(2 + 2) = 1/4. Students often struggle with conjugate multiplication or make algebraic errors. The systematic approach for radical limits is to rationalize using conjugates and simplify step by step.

This limit creates 0/0, requiring expansion of the cubic expression in the numerator. Expand (1+x)³ using the binomial theorem: (1+x)³ = 1 + 3x + 3x² + x³. The numerator becomes 1 + 3x + 3x² + x³ - 1 = 3x + 3x² + x³ = x(3 + 3x + x²). The expression becomes x(3 + 3x + x²)/x, which simplifies to (3 + 3x + x²) after canceling x. Substituting x = 0 gives 3 + 0 + 0 = 3. A common error is incorrectly expanding the cubic or not factoring out x properly. The technique is to expand binomial expressions, factor common terms, cancel, and substitute.

This limit involves a perfect square expression that creates 0/0, requiring careful algebraic simplification. The numerator (x-1)² can be written as (x-1)(x-1). The expression becomes (x-1)(x-1)/(x-1), which simplifies to (x-1) after canceling one factor of (x-1). Substituting x = 1 gives 1 - 1 = 0. Students often incorrectly cancel both factors or fail to recognize the perfect square structure. The key strategy is to expand or factor perfect squares appropriately, cancel only one instance of common factors, and substitute carefully.

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.

This limit initially gives $\frac{0}{0}$, requiring factorization of the numerator. The quadratic $x^2 + x - 2$ factors as $(x + 2)(x - 1)$ by finding two numbers that multiply to $-2$ and add to $1$. The expression becomes $\frac{(x + 2)(x - 1)}{x - 1}$, which simplifies to $(x + 2)$ after canceling the common factor $(x - 1)$. Substituting $x = 1$ gives $1 + 2 = 3$. Students often struggle with factoring quadratics or make sign errors. The key approach is to factor the numerator completely, identify common factors with the denominator, cancel appropriately, and substitute.

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.