This limit requires algebraic manipulation to resolve the indeterminate form 0/0 that occurs when directly substituting x = 3. The numerator $x^2 - 9$ factors as $(x - 3)(x + 3)$, allowing us to cancel the common factor $(x - 3)$ with the denominator. After cancellation, we get $\lim_{x \to 3} (x + 3) = 3 + 3 = 6$. A common error is to think the limit doesn't exist because of the 0/0 form, but factoring reveals the removable discontinuity. When facing rational functions with indeterminate forms, always try factoring first to simplify the expression.

Algebraic manipulation is essential for evaluating limits of rational functions that result in indeterminate forms like 0/0. To find the limit as x approaches 3 of (x² - 9)/(x - 3), factor the numerator as (x - 3)(x + 3). Cancel the common factor (x - 3) from numerator and denominator, simplifying to x + 3 for x ≠ 3. Substituting x = 3 into the simplified expression gives 6. A tempting distractor might be to plug in x = 3 directly without simplifying, yielding 0/0, which is indeterminate and could lead to incorrectly choosing DNE. Always factor and simplify rational expressions before taking the limit to resolve indeterminate forms.

This limit requires algebraic manipulation through factoring a difference of fourth powers. The expression (x⁴-1)/(x-1) gives 0/0 when x=1, so we need to factor the numerator. We can factor x⁴-1 as (x²-1)(x²+1), and then factor x²-1 further as (x-1)(x+1), giving us (x-1)(x+1)(x²+1). The full expression becomes [(x-1)(x+1)(x²+1)]/(x-1), and after canceling (x-1), we get (x+1)(x²+1). Substituting x=1 gives (1+1)(1²+1) = 2×2 = 4. A common error is to only partially factor x⁴-1 or to make arithmetic mistakes when evaluating the simplified expression. When dealing with xⁿ-1, remember it always has (x-1) as a factor, which often helps resolve 0/0 indeterminate forms.

This limit involves algebraic manipulation through rationalization to eliminate the indeterminate form 0/0. Multiplying both numerator and denominator by the conjugate (√(4+h) + 2) gives us [(√(4+h) - 2)(√(4+h) + 2)]/[h(√(4+h) + 2)] = [(4+h) - 4]/[h(√(4+h) + 2)] = h/[h(√(4+h) + 2)]. After canceling h, we get 1/(√(4+h) + 2), which evaluates to 1/(√4 + 2) = 1/4 as h approaches 0. Students often forget to multiply both parts of the fraction by the conjugate, leading to incorrect simplification. When square roots create indeterminate forms, rationalization is your primary algebraic tool.

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.

Algebraic manipulation with rationalization is vital for limits with square roots in indeterminate forms. To compute the limit as x approaches 1 of (√(3x + 1) - 2)/(x - 1), multiply by the conjugate √(3x + 1) + 2. This results in (3x + 1 - 4)/((x - 1)(√(3x + 1) + 2)) = 3(x - 1)/((x - 1)(√(3x + 1) + 2)), simplifying to 3/(√(3x + 1) + 2). Substituting x = 1 gives 3/4. Direct substitution yields 0/0, which might tempt choosing DNE, but rationalization clarifies the value. Rationalize numerators involving square roots to simplify and find limits in optics or similar models.

Determining limits using algebraic manipulation involves simplifying expressions to resolve indeterminate forms like $0/0$. To compute $\lim_{x\to1}\frac{x^3-1}{x-1}$, factor the numerator as $(x-1)(x^2 + x + 1)$ using the difference of cubes. Cancel the $(x-1)$ term, leaving $x^2 + x + 1$. As x approaches 1, this evaluates to 3. A tempting distractor like 0 might come from direct substitution without factoring, resulting in $0/0$. Always recognize polynomial factorizations like difference of powers and simplify before taking the limit.

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.