The value of `lim_(xrarr1)Sigma_(r=1)^(10)=(x^(r )-1^(r ))/(2(x-1))` is equal to

Answer

To solve the limit problem, we need to evaluate the expression:

\[ \lim_{x \to 1} \sum_{r=1}^{10} \frac{x^r - 1^r}{2(x - 1)} \]

### Step-by-step Solution:

1. **Rewrite the Expression**:
The term \(1^r\) is simply 1 for any integer \(r\). Therefore, we can rewrite the expression as:
\[ \lim_{x \to 1} \sum_{r=1}^{10} \frac{x^r - 1}{2(x - 1)} \]

2. **Factor the Numerator**:
The expression \(x^r - 1\) can be factored using the difference of powers:
\[ x^r - 1 = (x - 1)(x^{r-1} + x^{r-2} + \ldots + 1) \]
Substituting this into our limit gives:
\[ \lim_{x \to 1} \sum_{r=1}^{10} \frac{(x - 1)(x^{r-1} + x^{r-2} + \ldots + 1)}{2(x - 1)} \]

3. **Cancel Out the Common Terms**:
The \((x - 1)\) terms in the numerator and denominator cancel out, leading to:
\[ \lim_{x \to 1} \sum_{r=1}^{10} \frac{x^{r-1} + x^{r-2} + \ldots + 1}{2} \]

4. **Evaluate the Limit**:
As \(x\) approaches 1, each \(x^k\) (where \(k\) is any non-negative integer) approaches 1. Thus, we have:
\[ \lim_{x \to 1} \sum_{r=1}^{10} \frac{1 + 1 + \ldots + 1}{2} = \sum_{r=1}^{10} \frac{r}{2} \]
where each term in the summation contributes 1.

5. **Calculate the Sum**:
The sum of the first \(n\) natural numbers is given by the formula:
\[ S_n = \frac{n(n + 1)}{2} \]
For \(n = 10\):
\[ S_{10} = \frac{10 \times 11}{2} = 55 \]

6. **Final Calculation**:
Now, we divide this sum by 2:
\[ \frac{55}{2} = 27.5 \]

Thus, the value of the limit is:

\[ \boxed{27.5} \]