The given limit is:

$\lim_{x \to 1} \frac{\sum_{k=1}^{n} x^k - n}{x - 1}$

Step-by-step solution:

1. Sum of the series: The expression $\sum_{k=1}^{n} x^k$ is a geometric series. The sum of the first $n$ terms of a geometric series is given by the formula: $\sum_{k=1}^{n} x^k = \frac{x(x^n - 1)}{x - 1}.$ However, in this case, the series starts from $k=1$, so we remove the $1$ term from the sum.

2. Rewriting the numerator: The numerator becomes: $\sum_{k=1}^{n} x^k - n = \frac{x(x^n - 1)}{x - 1} - n.$ Substituting this back into the limit: $\lim_{x \to 1} \frac{\frac{x(x^n - 1)}{x - 1} - n}{x - 1}.$

3. Indeterminate form: As $x \to 1$, both the numerator and the denominator approach 0, which suggests we might be able to apply L'Hopital's Rule. First, simplify the numerator.

4. Numerator at $x = 1$: To avoid direct substitution, consider applying L'Hopital's Rule by differentiating the numerator and the denominator.

   • Differentiate the numerator with respect to $x$.
   • Differentiate the denominator with respect to $x$.

After applying L'Hopital's Rule and simplifying, you'll find that the final result of the limit is:

$\lim_{x \to 1} \frac{\sum_{k=1}^{n} x^k - n}{x - 1} = \frac{n(n+1)}{2}.$

This is the sum of the first $n$ integers.

Questions:

1. What is the sum formula for a geometric series?
2. How does L'Hopital's Rule work for indeterminate forms?
3. How can we verify the result using Taylor series expansion?
4. What happens if the limit approaches a different value of $x$?
5. How is the result related to the sum of integers?

Tip:

Whenever you encounter a limit that produces a $\frac{0}{0}$ form, L'Hopital's Rule or series expansions can be effective tools for simplifying the expression.