This rational function is continuous at $x = 3$ since $3 - 1 = 2 \neq 0$. Using the quotient rule: $\lim_{x\to 3} k(x) = \frac{\lim_{x\to 3} 5}{\lim_{x\to 3}(x - 1)} = \frac{5}{3 - 1} = \frac{5}{2}$. The numerator is constant, and the denominator approaches 2, so the limit is well-defined. A common error would be not checking that the denominator is non-zero or making arithmetic mistakes. The key strategy is to verify continuity by ensuring the denominator limit is non-zero, then apply direct substitution to both numerator and denominator.

This square root function is continuous at $x = 0$ since $25 + 4(0) = 25 > 0$. Using direct substitution: $\lim_{x\to 0} M(x) = \sqrt{25 + 4(0)} = \sqrt{25 + 0} = \sqrt{25} = 5$. The expression under the radical is positive, ensuring the function is well-defined. A common mistake would be arithmetic errors in $4(0) = 0$ or $\sqrt{25} = 5$. The key strategy is to verify that expressions under radicals remain non-negative, then apply direct substitution with careful arithmetic.

This linear function is continuous everywhere, allowing direct substitution. Using limit laws: $\lim_{x\to -1} f(x) = \frac{7}{2} - (-1) = \frac{7}{2} + 1 = \frac{7}{2} + \frac{2}{2} = \frac{9}{2}$. We apply the difference rule, carefully handling the double negative: $-(-1) = +1$. A common error would be sign mistakes with the double negative or incorrect fraction addition. The key strategy is to work systematically with signs in linear functions and use proper fraction arithmetic when combining terms.

This rational function with a quadratic numerator requires careful application of limit laws. Since the denominator is non-zero at $x = -2$, we can use direct substitution: $\lim_{x\to -2} f(x) = \frac{(-2)^2 + 1}{3} = \frac{4 + 1}{3} = \frac{5}{3}$. We apply the quotient rule, with the numerator limit being $(-2)^2 + 1 = 5$ and denominator limit being 3. A common error would be sign mistakes when substituting negative values or incorrectly computing $(-2)^2$. The key strategy is to carefully substitute negative values and apply order of operations correctly in rational functions.

This function combines a rational term with a constant. Using the sum rule: $\lim_{x\to 5} p(x) = \lim_{x\to 5} \frac{1}{x} + \lim_{x\to 5} 2 = \frac{1}{5} + 2 = \frac{1}{5} + \frac{10}{5} = \frac{11}{5}$. The rational function $\frac{1}{x}$ is continuous at $x = 5$ since the denominator is non-zero. A common mistake would be incorrectly adding the fractions or not finding a common denominator. The transferable approach is to evaluate each term separately using limit laws, then combine the results using appropriate arithmetic operations.

This quadratic function with a fractional coefficient is continuous at $x = 2$. Using direct substitution: $\lim_{x\to 2} f(x) = \frac{1}{2}(2)^2 + 4 = \frac{1}{2}(4) + 4 = 2 + 4 = 8$. We apply the sum rule and scalar multiplication rule, carefully computing $\frac{1}{2} \cdot 4 = 2$. A common mistake would be incorrectly handling the fractional coefficient or making arithmetic errors. The transferable strategy is to work step-by-step with fractional coefficients, ensuring proper order of operations in polynomial functions.

This linear function with fractional coefficients requires careful arithmetic. Using direct substitution: $\lim_{x\to 3} f(x) = \frac{2}{3}(3) + \frac{1}{3} = \frac{6}{3} + \frac{1}{3} = 2 + \frac{1}{3} = \frac{6}{3} + \frac{1}{3} = \frac{7}{3}$. We apply the sum rule and scalar multiplication rule, being careful with fraction arithmetic. A common mistake would be incorrectly computing $\frac{2}{3} \cdot 3 = 2$ or making errors when adding fractions. The transferable approach is to work systematically with fractional coefficients, ensuring proper arithmetic at each step.

This linear function with a fractional coefficient is continuous everywhere. Using direct substitution: $\lim_{x\to 6} r(x) = 7 + \frac{6}{3} = 7 + 2 = 9$. We apply the sum rule for limits, evaluating each term separately: the constant term gives 7, and the rational term gives $\frac{6}{3} = 2$. A common error would be arithmetic mistakes in the fraction division or not recognizing this as a continuous function. The key strategy is to identify continuous functions and apply direct substitution, treating each term according to the appropriate limit law.

This function involves the cosine function, which is continuous everywhere. Using limit laws and direct substitution: $\lim_{x\to 0} h(x) = 5 - 3\cos(0) = 5 - 3(1) = 5 - 3 = 2$. The key insight is that $\cos(0) = 1$, and we apply the difference rule and scalar multiplication rule for limits. A common mistake would be forgetting that $\cos(0) = 1$ or incorrectly applying the arithmetic operations. The transferable strategy is to memorize key trigonometric values and recognize that trigonometric functions are continuous at standard points, allowing direct substitution.

This linear function allows direct substitution since it's continuous everywhere. Using limit laws, $\lim_{t\to 3} s(t) = s(3)$ by the continuity of linear functions. Applying the sum rule and scalar multiplication rule: $s(3) = 4(3) - 7 = 12 - 7 = 5$. A common mistake would be misapplying the arithmetic or not recognizing that linear functions are continuous at all points. The key strategy is to identify continuous functions and use direct substitution, which works for all polynomial functions including linear ones.