This limit requires algebraic manipulation by rewriting to use the standard limit lim[u→0] sin(u)/u = 1. The expression sin(3x)/x can be rewritten by multiplying and dividing by 3: [3·sin(3x)]/(3x). This gives us 3·[sin(3x)/(3x)], and if we let u=3x, then as x→0, u→0 as well. The expression becomes 3·[sin(u)/u], and using the standard limit, we get 3×1 = 3. Students might incorrectly think the answer is 1 by forgetting to account for the coefficient 3, or they might choose 1/3 by incorrectly placing the 3 in the denominator. When you see sin(kx)/x, always rewrite it as k·[sin(kx)/(kx)] to apply the standard sine limit formula.

Determining limits using algebraic manipulation involves simplifying expressions to resolve indeterminate forms like $0/0$. For $f(x) = \frac{x^2-1}{x-1}$ where $x \ne 1$, factor the numerator as $(x-1)(x+1)$. Cancel the $(x-1)$ term, simplifying to $x+1$ for $x \ne 1$. As $x$ approaches 1, this equals 2. A tempting distractor like DNE might come from thinking the function is undefined at $x=1$, but the limit exists regardless. Always simplify the expression to find the limit even if the function has a removable discontinuity.

Determining limits using algebraic manipulation involves simplifying expressions to resolve indeterminate forms like 0/0. For \lim_${x\to3}$\frac{x^2$-9}{x-3}$, factor the numerator as (x-3)(x+3). Cancel the common (x-3) factor, yielding x+3. As x approaches 3, this equals 6. A tempting distractor like 0 might come from plugging in x=3 directly without simplifying, resulting in 0/0, which is indeterminate. Always factor and cancel common terms before taking the limit to avoid indeterminate forms.

This limit requires algebraic manipulation to resolve the indeterminate form 0/0. The expression (t²-9)/(t-3) can be factored by recognizing that t²-9 = (t+3)(t-3), which is a difference of squares. After factoring, we get [(t+3)(t-3)]/(t-3), and the (t-3) terms cancel, leaving us with t+3. Now we can evaluate the limit by direct substitution: lim[t→3] (t+3) = 3+3 = 6. A common error would be to substitute t=3 directly into the original expression without simplifying first, which would give 0/0 and lead to incorrectly choosing DNE. When faced with a rational function that gives 0/0, always try factoring first to cancel common factors before evaluating the limit.

This limit requires algebraic manipulation using the trigonometric identity 1 - cos(x) = 2sin²(x/2) to resolve the 0/0 form. Substituting this identity gives us lim[x→0] 2sin²(x/2)/x² = lim[x→0] 2 · [sin(x/2)/(x/2)]² · (1/4) = 2 · 1² · (1/4) = 1/2. The key insight is recognizing that sin(x/2)/(x/2) approaches 1 as x approaches 0, and we need the factor of 1/4 to adjust for the argument. Students often forget the adjustment factor when using half-angle formulas. For limits involving 1 - cos(x), the half-angle identity is your most powerful algebraic tool.

This limit requires algebraic manipulation using the factorization of x³ - 1 to eliminate the 0/0 form. The expression x³ - 1 factors as (x - 1)(x² + x + 1), allowing us to cancel the (x - 1) term with the denominator. After simplification, we get lim[x→1] (x² + x + 1) = 1² + 1 + 1 = 3. Students might try to use L'Hôpital's rule, but the problem specifically asks for algebraic manipulation. Remember that aⁿ - bⁿ can be factored as (a - b) times a sum of powers, which is crucial for solving such limits algebraically.

This limit requires algebraic manipulation by factoring both numerator and denominator to simplify the rational expression. The denominator x² - 4 factors as (x - 2)(x + 2), so we have lim[x→2] [(x-2)(x+5)]/[(x-2)(x+2)]. Canceling the common factor (x - 2) gives us lim[x→2] (x+5)/(x+2) = 7/4. Students might incorrectly cancel before checking if the factors are truly common throughout the domain. Always factor completely before canceling to avoid errors in limit evaluation.

This limit uses algebraic manipulation through the binomial expansion or recognizing it as a derivative definition to resolve 0/0. Using the binomial theorem, (1+x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵, so (1+x)⁵ - 1 = 5x + 10x² + 10x³ + 5x⁴ + x⁵. Dividing by x gives 5 + 10x + 10x² + 5x³ + x⁴, which approaches 5 as x approaches 0. A common mistake is trying to factor without recognizing the binomial pattern. When you see expressions like (1+x)ⁿ - 1, think binomial expansion or derivative definition for efficient algebraic manipulation.

This limit requires algebraic manipulation by rationalizing the denominator containing a square root. Multiply numerator and denominator by the conjugate √(x+6) + 3 to get [(x-3)(√(x+6) + 3)]/[(x+6) - 9] = [(x-3)(√(x+6) + 3)]/(x-3). Cancel the common factor (x-3) to obtain √(x+6) + 3. As x approaches 3, this becomes √(3+6) + 3 = √9 + 3 = 3 + 3 = 6. Students often rationalize incorrectly and get 1/6 by inverting the final answer. When the square root is in the denominator, multiply by its conjugate and simplify carefully.

This limit requires algebraic manipulation through factoring to resolve the indeterminate form 0/0. The numerator $x^2-2x$ can be factored as $x(x-2)$, giving us $\frac{x(x-2)}{x-2}$. Since we're finding the limit as $x$ approaches 2 (not evaluating at $x=2$), we can cancel the common factor $(x-2)$, leaving $\lim_{x\to 2} x$. Substituting $x=2$ into this simplified expression yields simply 2. A student might make an error by factoring as $2(x-2)$ or by thinking the answer should be 4 from incorrectly evaluating $x^2$ at $x=2$. When factoring expressions for limit problems, always factor completely and verify by expanding back before canceling common terms.