lim_x-1[∑_r=1²⁰⁰⁷x^r-2007]/x-1

To evaluate the limit you’ve presented, we can start by rewriting it in a more manageable form. The expression you have is:

lim (x → 1) [ ∑ (r=1 to 2007) x^r - 2007 ] / (x - 1).

This limit can be approached using L'Hôpital's Rule, which is useful when we encounter an indeterminate form like 0/0. First, let’s analyze the numerator:

Understanding the Summation

The summation ∑ (r=1 to 2007) x^r can be simplified. This is a geometric series where the first term is x and the common ratio is also x. The formula for the sum of a geometric series is:

S_n = a * (1 - r^n) / (1 - r),

where:

• a is the first term,
• r is the common ratio,
• n is the number of terms.

In our case, a = x, r = x, and n = 2007. Thus, we can express the sum as:

∑ (r=1 to 2007) x^r = x * (1 - x^2007) / (1 - x).

Substituting Back into the Limit

Now, substituting this back into our limit gives us:

lim (x → 1) [ (x * (1 - x^2007) / (1 - x)) - 2007 ] / (x - 1).

To simplify this further, we can combine the terms in the numerator:

lim (x → 1) [ (x * (1 - x^2007) - 2007(1 - x)) / (1 - x) ].

Evaluating the Limit

As x approaches 1, both the numerator and denominator approach 0, leading us to apply L'Hôpital's Rule. This involves differentiating the numerator and the denominator:

1. Differentiate the numerator:

Let f(x) = x * (1 - x^2007) - 2007(1 - x).

Using the product rule and chain rule, we find:

f'(x) = (1 - x^2007) + x * (-2007x^2006) + 2007.

2. Differentiate the denominator:

The derivative of (1 - x) is -1.

Now, we can rewrite our limit as:

lim (x → 1) f'(x) / -1.

Calculating the Derivative at x = 1

Substituting x = 1 into f'(x):

f'(1) = (1 - 1) + 1 * (-2007 * 1^2006) + 2007 = 0 - 2007 + 2007 = 0.

Since we still have an indeterminate form, we apply L'Hôpital's Rule again. We differentiate f'(x) again:

f''(x) = -2007 * 2006 * x^2005.

Now, substituting x = 1 into f''(x):

f''(1) = -2007 * 2006 * 1^2005 = -2007 * 2006.

Thus, our limit becomes:

lim (x → 1) f''(x) / -1 = 2007 * 2006.

Final Result

Therefore, the limit evaluates to:

2007 * 2006.

This result shows how the limit can be approached through careful manipulation of the summation and the application of calculus techniques. If you have any further questions about this process or any specific steps, feel free to ask!