Evaluate the following limits, if they exist :
`lim_(x to -1)[1+x+x^(2)+……+x^(10)]`.

Answer

To evaluate the limit \( \lim_{x \to -1} [1 + x + x^2 + \ldots + x^{10}] \), we can follow these steps:

### Step 1: Recognize the series
The expression \( 1 + x + x^2 + \ldots + x^{10} \) is a geometric series. The sum of a geometric series can be calculated using the formula:
\[ S_n = \frac{a(1 - r^{n+1})}{1 - r} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms minus one.

### Step 2: Identify parameters of the series
In our case:
- The first term \( a = 1 \)
- The common ratio \( r = x \)
- The number of terms is \( 11 \) (from \( x^0 \) to \( x^{10} \)), so \( n = 10 \)

### Step 3: Write the sum using the formula
Using the formula for the sum of a geometric series:
\[ S = \frac{1(1 - x^{11})}{1 - x} = \frac{1 - x^{11}}{1 - x} \]

### Step 4: Substitute \( x = -1 \) into the sum
Now we need to evaluate the limit as \( x \) approaches \(-1\):
\[ \lim_{x \to -1} \frac{1 - x^{11}}{1 - x} \]

### Step 5: Substitute directly
Substituting \( x = -1 \):
\[ \frac{1 - (-1)^{11}}{1 - (-1)} = \frac{1 - (-1)}{1 + 1} = \frac{1 + 1}{2} = \frac{2}{2} = 1 \]

### Conclusion
Thus, the limit exists and is equal to:
\[ \lim_{x \to -1} [1 + x + x^2 + \ldots + x^{10}] = 1 \]

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