The demand function is given as:

$p(x) = \frac{13,000}{1 - \frac{7}{7 + e^{-0.001x}}}$

Let's solve for the price $p$ when $x = 1000$, $x = 1500$, and determine the limit as $x \to \infty$.

(a) For $x = 1000$:

We substitute $x = 1000$ into the formula and calculate the corresponding price.

$p(1000) = \frac{13,000}{1 - \frac{7}{7 + e^{-0.001 \cdot 1000}}}$

(b) For $x = 1500$:

Substitute $x = 1500$ into the formula.

$p(1500) = \frac{13,000}{1 - \frac{7}{7 + e^{-0.001 \cdot 1500}}}$

(c) Limit of the price as $x \to \infty$:

As $x \to \infty$, $e^{-0.001x} \to 0$. Thus, the demand function simplifies to:

$p(x) \to \frac{13,000}{1 - \frac{7}{7}} = \frac{13,000}{1 - 1} = \infty$

So, the price increases without bound as $x \to \infty$.

Now, I will calculate the values for parts (a) and (b).The calculated prices are:

(a) For $x = 1000$ units, the price is approximately $260,363.65$ dollars.

(b) For $x = 1500$ units, the price is approximately $420,833.71$ dollars.

(c) As $x \to \infty$, the price $p(x) \to \infty$ as well, meaning the price increases without bound.

Let me know if you would like more details or have any questions.

Here are some related questions:

1. What happens to the price if the rate constant in the exponential changes?
2. How does the price behave for small values of $x$?
3. Can you explain how the exponential term affects the price growth?
4. What if the coefficient in front of $e^{-0.001x}$ were different, how would it impact $p(x)$?
5. How would the demand function change if a different constant was used instead of 13,000?

Tip: Always verify exponential terms in demand functions carefully—they control the rapidity of price changes with demand.