`lim_(xrarroo) (sum_(r=1)^(10)(x+r)^(2010))/((x^(1006)+1)(2x^(1004)+1))=`

Options

A. 5
B. 2010
C. `(502)/(1005)`
D. 0

Explanation

To solve the limit

\[ \lim_{x \to \infty} \frac{\sum_{r=1}^{10} (x+r)^{2010}}{(x^{1006}+1)(2x^{1004}+1)}, \]

we will break it down step by step.

### Step 1: Expand the numerator

The numerator is a summation of the form:

\[ \sum_{r=1}^{10} (x+r)^{2010}. \]

As \(x\) approaches infinity, we can approximate \((x+r)^{2010}\) using the binomial expansion:

\[ (x+r)^{2010} = x^{2010} \left(1 + \frac{r}{x}\right)^{2010}. \]

Using the binomial theorem, we have:

\[ \left(1 + \frac{r}{x}\right)^{2010} \approx 1 + \frac{2010r}{x} + O\left(\frac{1}{x^2}\right) \text{ as } x \to \infty. \]

Thus,

\[ (x+r)^{2010} \approx x^{2010} \left(1 + \frac{2010r}{x}\right) = x^{2010} + 2010r x^{2009} + O\left(x^{2008}\right). \]

Now, summing from \(r=1\) to \(10\):

\[ \sum_{r=1}^{10} (x+r)^{2010} \approx \sum_{r=1}^{10} \left(x^{2010} + 2010r x^{2009}\right) = 10x^{2010} + 2010 \sum_{r=1}^{10} r x^{2009}. \]

Calculating \(\sum_{r=1}^{10} r = \frac{10(10+1)}{2} = 55\):

\[ \sum_{r=1}^{10} (x+r)^{2010} \approx 10x^{2010} + 2010 \cdot 55 x^{2009} = 10x^{2010} + 110550 x^{2009}. \]

### Step 2: Simplify the denominator

The denominator is:

\[ (x^{1006}+1)(2x^{1004}+1). \]

As \(x\) approaches infinity, we can approximate this as:

\[ (x^{1006})(2x^{1004}) = 2x^{2010}. \]

### Step 3: Combine the results

Now we can substitute the approximations back into the limit:

\[ \lim_{x \to \infty} \frac{10x^{2010} + 110550 x^{2009}}{2x^{2010}}. \]

### Step 4: Divide by the highest power of \(x\)

We can factor out \(x^{2010}\) from the numerator:

\[ = \lim_{x \to \infty} \frac{x^{2010} \left(10 + \frac{110550}{x}\right)}{2x^{2010}}. \]

Cancelling \(x^{2010}\):

\[ = \lim_{x \to \infty} \frac{10 + \frac{110550}{x}}{2}. \]

As \(x \to \infty\), \(\frac{110550}{x} \to 0\):

\[ = \frac{10 + 0}{2} = \frac{10}{2} = 5. \]

### Final Answer

Thus, the limit is:

\[ \boxed{5}. \]