This problem requires using the limit laws for rational functions. The denominator at x=5 is 5 ≠0, so direct substitution is applicable. The limit is $(5^2$ -4*5 +1)/5 = (25-20+1)/5 =6/5. The choice A is this expression. A tempting distractor is D d$\frac{25-4(5)+1}{0}$, perhaps from incorrectly using 0 in denominator. A transferable strategy is to substitute the value into both numerator and denominator separately and then divide.

This problem requires using the limit laws, specifically the direct substitution property for polynomials. Since S(t) is a polynomial, it is continuous, and the limit as t approaches 2 is S(2) = $3(2)^2$ -5(2) +4. This expression simplifies to 12 -10 +4 =6. The choice D is this exact expression for direct substitution. A tempting distractor is C, which uses 6 as the coefficient instead of 3, perhaps from misreading $3t^2$ as $6t^2$, leading to an incorrect calculation. A transferable strategy is to identify if the function is continuous at the point and substitute directly for the limit.

This problem requires using limit laws on a rational function where the denominator is non-zero at the limit point. Since $0^2+1 = 1 \neq 0$, we can apply direct substitution: $\lim_{x\to 0} \frac{5-2x}{x^2+1} = \frac{5-2(0)}{0^2+1} = \frac{5-0}{0+1} = \frac{5}{1} = 5$. The limit laws allow us to evaluate the numerator and denominator separately when the denominator limit is non-zero. A potential error would be to think the limit is 0 because $x$ approaches 0, forgetting that we need to evaluate the entire expression. For rational functions, always substitute the limit value into both numerator and denominator, checking that the denominator remains non-zero.

This problem requires using the algebraic properties of limits, including simplifying rational expressions before taking the limit. The function R(x) = (x+1)(x-3)/(x-3) simplifies to x+1 for x≠3. Therefore, the limit as x approaches 3 is 3+1=4. The choice B is this value. A tempting distractor is A, which plugs into the original to get 0/0, an indeterminate form that doesn't give the limit. A transferable strategy is to factor and cancel common factors in rational functions when direct substitution gives 0/0.

This problem uses limit laws to evaluate a function involving subtraction of a rational term. Since $x=6$ makes the denominator non-zero, we can use direct substitution: $\lim_{x\to 6} \left(7-\frac{3}{x}\right) = 7-\frac{3}{6} = 7-\frac{1}{2} = \frac{14}{2}-\frac{1}{2} = \frac{13}{2}$. The limit laws tell us we can evaluate the limit of a difference by taking the difference of the individual limits. A potential mistake would be to compute $7-3 = 4$ and then divide by 6, getting $\frac{4}{6} = \frac{2}{3}$, by misapplying the order of operations. For expressions with multiple operations, always follow the correct order: division before subtraction.

This problem uses limit laws to evaluate a rational function where direct substitution is valid. Since the denominator $t+1$ is non-zero when $t=3$, we can substitute directly: $\lim_{t\to 3} \frac{t^2+1}{t+1} = \frac{3^2+1}{3+1} = \frac{9+1}{4} = \frac{10}{4} = \frac{5}{2}$. The limit laws allow us to evaluate the numerator and denominator limits separately when the denominator limit is non-zero. A common error would be to simplify incorrectly, perhaps getting $\frac{10}{3}$ by mistakenly using $t$ instead of $t+1$ in the denominator. For rational functions, always verify the denominator is non-zero at the limit point before applying direct substitution.

This problem requires using the limit laws for sums and quotients. The function is continuous at x=1 since denominator 1+5=6 ≠0. The limit is [3(1)-6]/(1+5) +2 = (-3)/6 +2 = -0.5 +2=1.5. The choice A is this expression. A tempting distractor is B d$\frac{3(1-6)}{1+5}$+2 = 3*(-5)/6 +2 = -15/6 +2 = -2.5+2=-0.5, from misplacing the parentheses in numerator. A transferable strategy is to apply substitution to each component and combine using the sum law.

This problem illustrates the use of limit laws, specifically the sum, difference, and constant multiple rules for limits. The expression $2x^2$ + 3x - 5 is a polynomial, so the limit as x approaches 1 is simply its value at x=1. Substituting x=1 gives $2*(1)^2$ + 3*1 - 5 = 2 + 3 - 5 = 0. This follows from applying the power rule to each term and then using the sum and difference properties of limits. A tempting distractor like -4 might come from confusing it with evaluating g(x), but the question asks only for the numerator, which is defined at x=1. A transferable strategy is to apply direct substitution for polynomials, as they are continuous everywhere.

This problem illustrates the use of limit laws, specifically the sum, difference, and constant multiple rules. m(x) is a polynomial, continuous everywhere, so the limit as x approaches 0 is m(0). Substituting x=0 gives (5/2)0 - 40 + 9 = 9. This follows from applying the power rule to each term and combining with sum properties. A tempting distractor like 0 might come from only considering the variable terms, but the constant 9 remains. A transferable strategy is to evaluate polynomials at the limit point directly due to their continuity.

This problem requires using the limit laws for rational functions. Since the denominator at t=1 is 1+3=4 ≠0, direct substitution is valid. The limit is $(1^2$ +3*1)/(1+3) = (1+3)/4 =1. The choice B is this plugged-in expression. A tempting distractor is D $d$\frac{1^2$ +3(1)}{3}$ =4/3, perhaps from misplacing the denominator as 3 alone. A transferable strategy is to evaluate the denominator's limit first to ensure it's not zero before applying direct substitution.