Question

$\lim_{x \to 1} \frac{\left(\sum_{k=1}^{100} x^{k}\right)-100}{x-1}$ is equal to

• A. 0
• B. 5050
• C. 4550
• D. -5050

Answer 1

Answer: (B)

5050

Answer 2

The given limit is $\lim_{x \to 1} \frac{\left(\sum_{k=1}^{100} x^{k}\right)-100}{x-1}$.

Let $f(x) = \sum_{k=1}^{100} x^{k} = x + x^2 + x^3 + \dots + x^{100}$.

We evaluate the numerator and denominator at $x=1$:

Numerator at $x=1$: $\left(\sum_{k=1}^{100} 1^{k}\right)-100 = \left(\sum_{k=1}^{100} 1\right)-100 = 100 - 100 = 0$.

Denominator at $x=1$: $1-1 = 0$.

Since the limit is of the indeterminate form $\frac{0}{0}$, we can use L'Hôpital's Rule or the definition of the derivative.

Method 1: Using the definition of the derivative.

Let $f(x) = \sum_{k=1}^{100} x^{k}$. Then $f(1) = \sum_{k=1}^{100} 1^{k} = 100$.

The limit can be written as $\lim_{x \to 1} \frac{f(x) - f(1)}{x-1}$. This is the definition of the derivative of $f(x)$ at $x=1$, i.e., $f'(1)$.

First, we find the derivative of $f(x)$:

$f'(x) = \frac{d}{dx} \left(\sum_{k=1}^{100} x^{k}\right) = \frac{d}{dx}(x + x^2 + \dots + x^{100})$.

Using the power rule $\frac{d}{dx}(x^k) = kx^{k-1}$, we differentiate each term:

$f'(x) = 1 \cdot x^{1-1} + 2 \cdot x^{2-1} + 3 \cdot x^{3-1} + \dots + 100 \cdot x^{100-1}$

$f'(x) = 1 + 2x + 3x^2 + \dots + 100x^{99}$.

Now, we evaluate $f'(1)$:

$f'(1) = 1 + 2(1) + 3(1)^2 + \dots + 100(1)^{99}$

$f'(1) = 1 + 2 + 3 + \dots + 100$.

This is the sum of the first 100 positive integers. The sum of the first $n$ positive integers is given by the formula $\frac{n(n+1)}{2}$.

For $n=100$, the sum is $\frac{100(100+1)}{2} = \frac{100 \times 101}{2} = 50 \times 101 = 5050$.

Therefore, the limit is 5050.

Method 2: Using L'Hôpital's Rule.

Let $N(x) = \left(\sum_{k=1}^{100} x^{k}\right)-100$ and $D(x) = x-1$.

Since $\lim_{x \to 1} N(x) = 0$ and $\lim_{x \to 1} D(x) = 0$, we can apply L'Hôpital's Rule:

$\lim_{x \to 1} \frac{N(x)}{D(x)} = \lim_{x \to 1} \frac{N'(x)}{D'(x)}$.

$N'(x) = \frac{d}{dx} \left(\sum_{k=1}^{100} x^{k}-100\right) = \frac{d}{dx}\left(\sum_{k=1}^{100} x^{k}\right) - \frac{d}{dx}(100)$

$N'(x) = \sum_{k=1}^{100} \frac{d}{dx}(x^{k}) - 0 = \sum_{k=1}^{100} kx^{k-1} = 1 + 2x + 3x^2 + \dots + 100x^{99}$.

$D'(x) = \frac{d}{dx}(x-1) = 1$.

So, the limit is $\lim_{x \to 1} \frac{1 + 2x + 3x^2 + \dots + 100x^{99}}{1}$.

Substituting $x=1$:

$1 + 2(1) + 3(1)^2 + \dots + 100(1)^{99} = 1 + 2 + 3 + \dots + 100$.

This is the sum of the first 100 positive integers, which is $\frac{100(100+1)}{2} = 5050$.

Both methods give the same result.