This exponential function is continuous everywhere, allowing direct substitution. Using limit laws: $\lim_{x\to 2} f(x) = 2^2 + 3 = 4 + 3 = 7$. We apply the sum rule, with the exponential term $2^2 = 4$ and the constant term 3. A common mistake would be confusing exponential and polynomial notation or making arithmetic errors. The transferable strategy is to recognize that exponential functions are continuous everywhere, so direct substitution applies, and to compute exponential values carefully.

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 rational function with a constant denominator is continuous at $x = 0$. Using direct substitution: $\lim_{x\to 0} p(x) = \frac{6 - 0}{3} = \frac{6}{3} = 2$. The function is linear in the numerator with a constant denominator, making it continuous everywhere. A common error would be arithmetic mistakes in $6 - 0 = 6$ or $\frac{6}{3} = 2$. The key strategy is to recognize that rational functions with constant non-zero denominators are continuous everywhere, allowing direct substitution.

This linear function with fractional coefficients is continuous everywhere. Using direct substitution: $\lim_{x\to 8} g(x) = \frac{8}{5} + \frac{2}{5} = \frac{8 + 2}{5} = \frac{10}{5} = 2$. We can also factor as $g(x) = \frac{x + 2}{5}$, giving $\frac{8 + 2}{5} = \frac{10}{5} = 2$. A common mistake would be not combining fractions properly or arithmetic errors. The transferable approach is to recognize when expressions can be simplified by factoring or by combining like terms with common denominators.

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 exponential function with a coefficient is continuous everywhere. Using direct substitution: $\lim_{t\to 0} N(t) = 100(1.02)^0 = 100(1) = 100$. We apply the property that any non-zero number raised to the power 0 equals 1, specifically $(1.02)^0 = 1$. A common error would be forgetting that $a^0 = 1$ for any $a \neq 0$ or making mistakes with the coefficient. The key strategy is to remember fundamental exponent rules and recognize that exponential functions are continuous, allowing direct substitution.

This logarithmic function is continuous at $x = 1$ since the argument is positive. Using direct substitution: $\lim_{x\to 1} g(x) = \ln(1) + 1 = 0 + 1 = 1$. We apply the sum rule, noting that $\ln(1) = 0$ by the definition of natural logarithm. A common error would be incorrectly recalling that $\ln(1) = 1$ instead of $\ln(1) = 0$, or not checking the domain of the logarithm. The key strategy is to memorize fundamental logarithmic values and verify that the argument is positive before applying direct substitution.

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 cubic function is continuous everywhere, allowing direct substitution. Using limit laws: $\lim_{t\to -1} T(t) = (-1)^3 - 2(-1) = -1 + 2 = 1$. We apply the difference rule and power rule, carefully handling the negative input: $(-1)^3 = -1$ and $-2(-1) = +2$. A common mistake would be sign errors when working with negative values, particularly $(-1)^3 = -1$ or the double negative in $-2(-1)$. The transferable strategy is to work methodically with signs when substituting negative values into polynomial expressions.

This rational function is continuous at $x = 2$ since $2 + 1 = 3 \neq 0$. Using the quotient rule: $\lim_{x\to 2} w(x) = \frac{\lim_{x\to 2}(3x)}{\lim_{x\to 2}(x + 1)} = \frac{3(2)}{2 + 1} = \frac{6}{3} = 2$. Both numerator and denominator limits exist, and the denominator is non-zero, allowing direct application of the quotient rule. A common mistake would be not checking that the denominator is non-zero or making arithmetic errors in the fraction simplification. The transferable approach is to verify continuity, then apply the quotient rule systematically.