The limit of a right Riemann sum is given by $$\lim_{n \to \infty} \sum_{k=1}^{n} f(a+k\Delta x) \Delta x = \int_a^b f(x) dx$$, where $$\Delta x = \frac{b-a}{n}$$. By matching the given expression, we can identify $$\Delta x = \frac{\pi}{n}$$ and $$f(x_k) = \sin(\frac{\pi k}{n})$$. Let's choose the starting point $$a=0$$. Then $$b-a = n \Delta x = n(\frac{\pi}{n}) = \pi$$, so $$b=\pi$$. The sample points are $$x_k = a+k\Delta x = 0 + k\frac{\pi}{n} = \frac{\pi k}{n}$$. The function is $$f(x) = \sin(x)$$. Thus, the definite integral is $$\int_0^\pi \sin(x) dx$$.

(B) is incorrect. The Riemann sum for this integral would be $$\lim_{n \to \infty} \sum_{k=1}^{n} \sin(\frac{\pi k}{n}) \frac{1}{n}$$, which is missing the factor of $$\pi$$ in the term corresponding to $$\Delta x$$.

(C) is incorrect because the function in the integrand is $$x \sin(x)$$, which does not match the sum.

(D) is incorrect because the interval of integration is $$[0,1]$$, which does not match the given sum.

The notation $$\lim_{x \to -3} g(x) = \infty$$ means that as $$x$$ gets closer and closer to -3 (from both the left and the right), the corresponding function values $$g(x)$$ increase without bound. This is accurately described by statement (A).

(B) is incorrect because it describes the limit $$\lim_{x \to \infty} g(x) = -3$$.

(C) is incorrect because infinity is not a number, and a limit describes the behavior approaching a point, not the value at the point. $$g(-3)$$ is likely undefined.

(D) is incorrect because it describes the limit $$\lim_{x \to \infty} g(x) = \infty$$.

A limit at infinity describes the end behavior of a function. The statement $$\lim_{x \to \infty} f(x) = L$$ means that the function values $$f(x)$$ get arbitrarily close to $$L$$ as $$x$$ increases without bound. Graphically, this corresponds to a horizontal asymptote at the line $$y=L$$. In this case, $$L=7$$.

(A) is incorrect; a vertical asymptote at $$x=7$$ would be represented by a limit like $$\lim_{x \to 7} f(x) = \pm\infty$$.

(C) is incorrect because a limit at infinity describes end behavior, not behavior at a specific point.

(D) is incorrect; an x-intercept at $$x=7$$ means $$f(7)=0$$, which is unrelated to the limit at infinity.

The correct answer translates the verbal description into limit notation. 'Approaches 2 from values less than 2' is represented by $$x \to 2^-$$. 'Values that get arbitrarily close to 5' is represented by the limit equaling 5. Therefore, the correct statement is $$\lim_{x \to 2^-} f(x) = 5$$.

(A) is incorrect because $$x \to 2^+$$ represents approaching 2 from values greater than 2.

(C) is incorrect because it reverses the roles of the input variable and the output value of the function.

(D) is incorrect because the concept of a limit describes the behavior of a function near a point, not necessarily at the point. The value of $$f(2)$$ could be different from 5 or even undefined.

The given limit matches the definition of the derivative, $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$. By comparing the terms, we can identify $$f(a+h) = e^{2(1+h)}$$ and $$f(a) = e^2$$. From $$f(a) = e^2$$, if we assume $$f(x)=e^{2x}$$, then $$e^{2a}=e^2$$, which implies $$a=1$$. We can verify: $$f(a+h) = f(1+h) = e^{2(1+h)}$$, which matches the numerator.

(A) is incorrect. If $$f(x)=e^x$$ and $$a=2$$, the limit would be $$\lim_{h \to 0} \frac{e^{2+h} - e^2}{h}$$.

(C) is incorrect. If $$f(x)=e^{x+2}$$ and $$a=0$$, then $$f(a) = e^2$$ but $$f(a+h) = e^{h+2} \neq e^{2(1+h)}$$.

(D) is incorrect. If $$f(x)=2e^x$$ and $$a=1$$, then $$f(a) = 2e \neq e^2$$.

The condition $$\lim_{x \to a} f(x) = 0$$ means that the function values of $$f$$ approach 0 as $$x$$ approaches $$a$$. If $$f$$ is also continuous at $$x=a$$, this implies that $$f(a)=0$$. Graphically, a point where a function's value is zero is an x-intercept. Therefore, for continuous functions, both graphs would have an x-intercept at $$x=a$$. This is the most direct graphical interpretation.

(A) is incorrect. Horizontal asymptotes relate to limits as $$x \to \pm\infty$$, not as $$x \to a$$.

(B) is only true if $$a=0$$. The statement must hold for any value of $$a$$.

(D) is not necessarily true. Having a root at $$x=a$$ does not guarantee a local extremum; for example, $$f(x)=(x-a)^3$$ has a root but no extremum at $$x=a$$.

The Squeeze Theorem states that if $$g(x) \le f(x) \le h(x)$$ for all $$x$$ in an open interval containing $$c$$ (except possibly at $$c$$ itself), and if $$\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$$, then $$\lim_{x \to c} f(x) = L$$. In this case, $$g(x)=-x^2$$, $$h(x)=x^2$$, and we find the limits of both bounding functions as $$x \to 0$$. Since both limits are 0, the limit of the function in the middle must also be 0. Option (A) correctly represents this logical step.

(B) is an incomplete justification; the Squeeze Theorem requires showing the limits of both bounding functions are equal.

(C) misunderstands the purpose of the theorem, which is specifically designed to handle cases involving oscillating functions like this one.

(D) is factually incorrect, as $$\lim_{x \to 0} (-x^2) = 0$$ and $$\lim_{x \to 0} x^2 = 0$$.

Connecting multiple representations of limits is a key skill in AP Calculus BC that requires interpreting tables, graphs, and algebraic forms to determine limit values. The table showing values approaching -1 from the left and 1 from the right indicates the limit does not exist as x approaches 0. The graph with a jump at x=0 represents different one-sided limits. Algebraically, r(x) = |x|/x is -1 for x<0 and 1 for x>0, confirming the limit does not exist. A tempting distractor might be choice A, which incorrectly claims the table approaches 1 from both sides, overlooking the absolute value's effect. To connect representations effectively, always check one-sided limits in functions with absolute values or discontinuities.

This problem requires connecting multiple representations of limits for s(x) = 1/(x-4). Algebraically, as x approaches 4 from the left, x-4 approaches 0 through negative values, making s(x) approach -∞; from the right, x-4 approaches 0 through positive values, making s(x) approach +∞. A table would show s(3.9) = -10, s(3.99) = -100, s(4.1) = 10, s(4.01) = 100, confirming the one-sided limits approach opposite infinities. Graphically, this creates a vertical asymptote at x=4 with the function going to -∞ on the left and +∞ on the right. Choice B incorrectly claims the limit is 0, perhaps thinking of horizontal asymptotes instead of vertical ones. When one-sided limits approach different infinities, the two-sided limit does not exist, which all three representations must consistently show.

The provided table of values shows that as $$t$$ approaches 1 from both the left (with values like 0.9 and 0.99) and the right (with values like 1.01 and 1.1), the corresponding function values $$h(t)$$ are getting closer and closer to 4. This numerical representation strongly suggests that the limit is 4.

(A) is incorrect because it is one of the function values for an input close to 1, but not the value being approached from both sides.

(C) is incorrect because 1 is the value that the input variable $$t$$ is approaching, not the value the function is approaching.

(D) is incorrect because the values from the left and the right are both approaching the same number, 4.