We need to evaluate the limit:
\(\lim_{x \to 1} \frac{\sum_{k=1}^{100} x^k - 100}{x - 1}\)
This expression can be interpreted using the formula for the sum of a geometric series. The series \(\sum_{k=1}^{n} x^k\) 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:
\(S_n = \frac{x(x^n - 1)}{x - 1}\)
For our problem, \(n = 100\), so we have:
\(\sum_{k=1}^{100} x^k = \frac{x(x^{100} - 1)}{x - 1}\)
Substituting this into the original expression, we get:
\(\lim_{x \to 1} \frac{\frac{x(x^{100} - 1)}{x - 1} - 100}{x - 1}\)
Simplifying the expression inside the limit:
\(= \frac{x(x^{100} - 1) - 100(x - 1)}{(x - 1)^2}\)
Now we evaluate this limit using L'Hopital's Rule, which is applicable because both the numerator and the denominator tend to 0 as \(x \to 1\). Taking derivatives of the numerator and the denominator:
\(\text{Numerator: } \frac{d}{dx}[x(x^{100} - 1) - 100(x - 1)] = \frac{d}{dx}[x^{101} - x - 100x + 100] = 101x^{100} - 101\) \(\text{Denominator: } \frac{d}{dx}[(x - 1)^2] = 2(x - 1)\)
Applying L'Hopital's Rule:
\(\lim_{x \to 1} \frac{101x^{100} - 101}{2(x - 1)}\)
We simplify further using L'Hopital's Rule again, as the new expression also gives a \(\frac{0}{0}\) form when evaluated at \(x = 1\).
\(\text{Numerator: } \frac{d}{dx}[101x^{100} - 101] = 10100x^{99}\) \(\text{Denominator: } \frac{d}{dx}[2(x - 1)] = 2\)
Now the limit becomes:
\(\lim_{x \to 1} \frac{10100x^{99}}{2} = \frac{10100 \times 1}{2} = 5050\)
Hence, the value of the limit is 5050.
Find \( P(0<X<5) \).