We need to evaluate the limit:

$\lim_{x \to 1} \frac{x^{1000} - 1}{x - 1}$

Step 1: Recognizing the Indeterminate Form

Substituting $x = 1$:

$\frac{1^{1000} - 1}{1 - 1} = \frac{0}{0}$

This is an indeterminate form, so we apply L'Hôpital's Rule, which states:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}, \quad \text{if the limit exists.}$

Step 2: Differentiation

We differentiate the numerator and denominator:

$\frac{d}{dx} (x^{1000} - 1) = 1000x^{999}, \quad \frac{d}{dx} (x - 1) = 1.$

Applying L'Hôpital's Rule:

$\lim_{x \to 1} \frac{1000x^{999}}{1}.$

Step 3: Evaluating at $x = 1$

$1000(1)^{999} = 1000.$

Conclusion

$\lim_{x \to 1} \frac{x^{1000} - 1}{x - 1} = 1000.$

Would you like a more detailed explanation?

Here are five related questions to explore:

1. How does L'Hôpital’s Rule help in evaluating indeterminate limits?
2. Can this limit be evaluated using algebraic simplification instead of L'Hôpital's Rule?
3. What is the general formula for $\lim_{x \to 1} \frac{x^n - 1}{x - 1}$ for any positive integer $n$?
4. How does the concept of the derivative relate to limits like this?
5. What happens if we change the exponent from 1000 to a general function $f(x)$?

Tip: The limit $\lim_{x \to a} \frac{x^n - a^n}{x - a}$ is always equal to $n a^{n-1}$, which is the derivative of $x^n$ at $x = a$.