Evaluate `lim_(x to 1) (sum_(k=1)^(100) x^(k) - 100)/(x-1).`

Explanation

To evaluate the limit

\[ \lim_{x \to 1} \frac{\sum_{k=1}^{100} x^k - 100}{x - 1}, \]

we can follow these steps:

### Step 1: Expand the summation
The summation \(\sum_{k=1}^{100} x^k\) can be expressed as:

\[ x + x^2 + x^3 + \ldots + x^{100}. \]

This is a geometric series with the first term \(a = x\) and common ratio \(r = x\). The formula for the sum of the first \(n\) terms of a geometric series is:

\[ S_n = a \frac{1 - r^n}{1 - r}. \]

In our case, \(n = 100\), so we have:

\[ \sum_{k=1}^{100} x^k = x \frac{1 - x^{100}}{1 - x}. \]

### Step 2: Substitute the summation into the limit
Now we substitute this back into the limit expression:

\[ \lim_{x \to 1} \frac{x \frac{1 - x^{100}}{1 - x} - 100}{x - 1}. \]

### Step 3: Simplify the expression
We can rewrite the expression in the limit:

\[ \lim_{x \to 1} \frac{x(1 - x^{100}) - 100(1 - x)}{(1 - x)(x - 1)}. \]

This simplifies to:

\[ \lim_{x \to 1} \frac{x - x^{101} - 100 + 100x}{(1 - x)(x - 1)} = \lim_{x \to 1} \frac{101x - x^{101} - 100}{(1 - x)(x - 1)}. \]

### Step 4: Evaluate the limit using L'Hôpital's Rule
As \(x\) approaches 1, both the numerator and denominator approach 0, so we can apply L'Hôpital's Rule:

\[ \lim_{x \to 1} \frac{d}{dx}(101x - x^{101} - 100) \Big/ \frac{d}{dx}((1 - x)(x - 1)). \]

Calculating the derivatives:

1. The derivative of the numerator \(101x - x^{101} - 100\) is \(101 - 101x^{100}\).
2. The derivative of the denominator \((1 - x)(x - 1)\) is \(-2(x - 1)\).

Thus, we have:

\[ \lim_{x \to 1} \frac{101 - 101x^{100}}{-2(x - 1)}. \]

### Step 5: Substitute \(x = 1\)
Now substituting \(x = 1\):

\[ \frac{101 - 101 \cdot 1^{100}}{-2(1 - 1)} = \frac{101 - 101}{0} = \text{indeterminate form}. \]

We need to apply L'Hôpital's Rule again:

Taking the derivative again gives:

1. The derivative of the numerator \(101 - 101x^{100}\) is \(-10100x^{99}\).
2. The derivative of the denominator \(-2(x - 1)\) is \(-2\).

Now we have:

\[ \lim_{x \to 1} \frac{-10100x^{99}}{-2} = \frac{10100}{2} = 5050. \]

### Final Answer
Thus, the limit evaluates to:

\[ \lim_{x \to 1} \frac{\sum_{k=1}^{100} x^k - 100}{x - 1} = 5050. \]

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