`lim_(xrarr-1)((x^4+x^2+x+1)/(x^2-x+1))^((1-cos(x+1))/((x+1)^2))` is equal to

Options

A. 1
B. `sqrt(2//3)`
C. `sqrt(3//2)`
D. `e^(1//2)`

Explanation

To solve the limit

\[ \lim_{x \to -1} \left( \frac{x^4 + x^2 + x + 1}{x^2 - x + 1} \right)^{\frac{1 - \cos(x + 1)}{(x + 1)^2}}, \]

we will follow these steps:

### Step 1: Evaluate the limit of the base expression
First, we need to evaluate the base of the limit as \( x \) approaches \(-1\):

\[ \frac{x^4 + x^2 + x + 1}{x^2 - x + 1}. \]

Substituting \( x = -1 \):

\[ \text{Numerator: } (-1)^4 + (-1)^2 + (-1) + 1 = 1 + 1 - 1 + 1 = 2, \]
\[ \text{Denominator: } (-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3. \]

Thus,

\[ \frac{x^4 + x^2 + x + 1}{x^2 - x + 1} \to \frac{2}{3} \text{ as } x \to -1. \]

### Step 2: Evaluate the limit of the exponent
Next, we need to evaluate the exponent:

\[ \frac{1 - \cos(x + 1)}{(x + 1)^2}. \]

Substituting \( x = -1 \):

\[ 1 - \cos(0) = 1 - 1 = 0, \]
\[ (x + 1)^2 = 0^2 = 0. \]

This gives us the indeterminate form \( \frac{0}{0} \). We can use L'Hôpital's Rule to evaluate this limit.

### Step 3: Apply L'Hôpital's Rule
Taking derivatives of the numerator and denominator:

1. The derivative of \( 1 - \cos(x + 1) \) is \( \sin(x + 1) \).
2. The derivative of \( (x + 1)^2 \) is \( 2(x + 1) \).

Now we have:

\[ \lim_{x \to -1} \frac{1 - \cos(x + 1)}{(x + 1)^2} = \lim_{x \to -1} \frac{\sin(x + 1)}{2(x + 1)}. \]

Substituting \( x = -1 \):

\[ \text{Numerator: } \sin(0) = 0, \]
\[ \text{Denominator: } 2(0) = 0. \]

Again, we have \( \frac{0}{0} \). We apply L'Hôpital's Rule again.

### Step 4: Apply L'Hôpital's Rule again
Taking derivatives again:

1. The derivative of \( \sin(x + 1) \) is \( \cos(x + 1) \).
2. The derivative of \( 2(x + 1) \) is \( 2 \).

Now we have:

\[ \lim_{x \to -1} \frac{\sin(x + 1)}{2(x + 1)} = \lim_{x \to -1} \frac{\cos(x + 1)}{2}. \]

Substituting \( x = -1 \):

\[ \cos(0) = 1. \]

Thus,

\[ \lim_{x \to -1} \frac{\sin(x + 1)}{2(x + 1)} = \frac{1}{2}. \]

### Step 5: Combine results
Now we can combine our results:

The limit becomes:

\[ \left( \frac{2}{3} \right)^{\frac{1}{2}} = \sqrt{\frac{2}{3}}. \]

### Final Answer
Thus, the final result is:

\[ \lim_{x \to -1} \left( \frac{x^4 + x^2 + x + 1}{x^2 - x + 1} \right)^{\frac{1 - \cos(x + 1)}{(x + 1)^2}} = \sqrt{\frac{2}{3}}. \]