Because f is continuous at $$\lim_{x \to 2} g(x) = 3$$, we can use the property for the limit of a composite function: $$\lim_{x \to 2} f(g(x)) = f(\lim_{x \to 2} g(x)) = f(3) = 5$$. Distractor B is the limit of the inner function, not the composite function. Distractor D is the sum of the input and output values ($$3+5$$).

Using the constant multiple, power, and product properties of limits: $$\lim_{x \to -2} (2[f(x)]^2 g(x)) = 2 \cdot(\lim_{x \to -2} f(x))^2 \cdot(\lim_{x \to -2} g(x)) = 2 \cdot(3)^2 \cdot(-1) = 2 \cdot 9 \cdot(-1) = -18$$. Distractor D results from ignoring the square on f(x) ($$2 \cdot 3 \cdot(-1) = -6$$). Distractor C is a sign error. Distractor B may result from incorrectly applying the power ($$2 \cdot(2 \cdot 3) \cdot(-1) = -12$$).

Let $$L = \lim_{x \to 1} f(x)$$ and $$M = \lim_{x \to 1} g(x)$$. The given information creates a system of equations: $$\frac{L}{M} = 3$$ and $$L+M=8$$. From the first equation, $$L = 3M$$. Substituting into the second gives $$3M + M = 8$$, which means $$4M=8$$ and $$M=2$$. Therefore, $$L = 3M = 3(2) = 6$$. Distractor A is the value for the limit of g(x). Distractor B is the value of the quotient. Distractor D would be the answer if the second equation was $$f(x)-g(x)=8$$.

If $$\lim_{x \to c} [f(x) + g(x)]$$ were to exist (call it M), then $$\lim_{x \to c} g(x) = \lim_{x \to c} ([f(x) + g(x)] - f(x)) = M - L$$, which would exist. This is a contradiction. Therefore, the limit of the sum must not exist. For A, the limit could exist if $$L=0$$. For B, the limit could exist if, for example, $$f(x)$$ was a large multiple of $$g(x)$$. For D, the limit could exist; for example, if $$g(x)$$ alternates between 1 and -1.

First, find the limit of the inner function: $$\lim_{x \to 1} (4f(x) - 5) = 4 \lim_{x \to 1} f(x) - 5 = 4(2) - 5 = 3$$. Since g is continuous at 3, we can evaluate the limit of the composite function: $$\lim_{x \to 1} g(4f(x) - 5) = g(\lim_{x \to 1} (4f(x) - 5)) = g(3) = -4$$. Distractor C is the limit of the inner function. Distractor B is the limit of f(x).

Let $$L = \lim_{x \to -2} g(x)$$. The quotient property states $$\frac{\lim_{x \to -2} f(x)}{\lim_{x \to -2} g(x)} = -3$$. Substituting the known value gives $$\frac{6}{L} = -3$$. Solving for L yields $$L = \frac{6}{-3} = -2$$. Distractor A is the result of multiplication instead of division. Distractor C is a sign error.

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.