To find the long-term behavior of the population, we need to evaluate the limit of $$P(t)$$ as $$t \to \infty$$. As $$t \to \infty$$, the term $$-0.2t$$ approaches $$-\infty$$, so $$e^{-0.2t}$$ approaches 0. The limit is then $$ \lim_{t \to \infty} \frac{2000}{1 + 49e^{-0.2t}} = \frac{2000}{1 + 49(0)} = 2000 $$. This value is the carrying capacity of the population.

The question asks for the long-term behavior of the concentration, which is found by evaluating the limit of $$C(t)$$ as $$t \to \infty$$. We evaluate $$ \lim_{t \to \infty} \frac{150t}{t^2 + 4} $$. Since the degree of the denominator (2) is greater than the degree of the numerator (1), the limit is 0. So, the long-term concentration of the drug approaches 0 mg/L.

To find the limit of this rational expression at infinity, we compare the leading terms of the numerator and the denominator. The leading term of the numerator is $$(2x)^3 = 8x^3$$. The leading term of the denominator is $$x(4x)^2 = x(16x^2) = 16x^3$$. The limit is the ratio of the coefficients of these leading terms, which is $$ \frac{8}{16} = \frac{1}{2} $$.

To find the limit of a rational function as $$x \to \infty$$, we compare the degrees of the numerator and the denominator. Since the degrees are the same (both are 2), the limit is the ratio of the leading coefficients. The leading coefficient of the numerator is 3, and the leading coefficient of the denominator is -2. Therefore, the limit is $$ \frac{3}{-2} = -\frac{3}{2} $$.

As $$x \to -\infty$$, the term $$e^x$$ approaches 0. Similarly, $$e^{2x} = (e^x)^2$$ also approaches 0. Substituting these values into the expression, we get $$ \lim_{x \to -\infty} \frac{e^x + 5}{e^{2x} - 3} = \frac{0 + 5}{0 - 3} = -\frac{5}{3} $$.

This question concerns the relative growth rates of functions. Exponential functions like $$e^x$$ grow faster than any polynomial function, including $$x^{100}$$. Therefore, as $$x \to \infty$$, the denominator grows much faster than the numerator, and the limit is 0. For B, the limit is $$ \lim_{x \to \infty} e^x = \infty $$. For C, the limit is $$ \lim_{x \to \infty} x \ln(x) = \infty $$. For D, the limit is $$ \lim_{x \to \infty} (\frac{5}{4})^x = \infty $$.

To evaluate this limit, we can divide the numerator and the denominator by the highest power of $$x$$, which is $$x$$. This gives $$ \lim_{x \to \infty} \frac{3 + \frac{\sin(x)}{x}}{2 - \frac{\cos(x)}{x}} $$. Since $$\sin(x)$$ and $$\cos(x)$$ are bounded between -1 and 1, $$ \lim_{x \to \infty} \frac{\sin(x)}{x} = 0 $$ and $$ \lim_{x \to \infty} \frac{\cos(x)}{x} = 0 $$ by the Squeeze Theorem. Therefore, the limit is $$ \frac{3+0}{2-0} = \frac{3}{2} $$.

The statement $$ \lim_{x \to \infty} f(x) = 4 $$ is the definition of a horizontal asymptote at $$y=4$$. Option B confuses horizontal and vertical asymptotes. Option C is not necessarily true; the limit as $$x \to -\infty$$ could be different or not exist. Option D is incorrect because a function can cross its horizontal asymptote.

To evaluate the limit as $$x \to -\infty$$, we divide the numerator and denominator by the highest power of $$x$$ in the denominator, which is $$ \sqrt{x^2} = |x| $$. Since $$x \to -\infty$$, $$x$$ is negative, so $$|x| = -x$$. Divide the numerator by $$x$$ and the denominator by $$-x = \sqrt{x^2}$$. $$ \lim_{x \to -\infty} \frac{(x+2)/x}{(\sqrt{9x^2-5x})/(-x)} = \lim_{x \to -\infty} \frac{1+2/x}{-\sqrt{(9x^2-5x)/x^2}} = \lim_{x \to -\infty} \frac{1+2/x}{-\sqrt{9-5/x}} = \frac{1+0}{-\sqrt{9-0}} = -\frac{1}{3} $$.

We can rewrite $$ \ln(x^2) $$ as $$2\ln(x)$$. The limit becomes $$ \lim_{x \to \infty} \frac{2\ln(x)}{3x} $$. Since polynomial functions (like $$3x$$) grow faster than logarithmic functions (like $$2\ln(x)$$), the limit is 0. Alternatively, applying L'Hôpital's Rule since the limit is of the form $$ \frac{\infty}{\infty} $$, we get $$ \lim_{x \to \infty} \frac{d/dx(2\ln(x))}{d/dx(3x)} = \lim_{x \to \infty} \frac{2/x}{3} = \lim_{x \to \infty} \frac{2}{3x} = 0 $$.