`lim_(x->1)[(x^3+2x^2+x+1)/(x^2+2x+3)]^((1-cos(x-1))/(x-1)^2)`

Options

A. e
B. `e^(1//2)`
C. 1
D. none of these

Answer

To solve the limit

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

we will follow these steps:

### Step 1: Evaluate the limit of the base as \( x \to 1 \)

First, we will substitute \( x = 1 \) into the expression:

\[ \text{Numerator: } 1^3 + 2 \cdot 1^2 + 1 + 1 = 1 + 2 + 1 + 1 = 5, \]
\[ \text{Denominator: } 1^2 + 2 \cdot 1 + 3 = 1 + 2 + 3 = 6. \]

Thus, the base evaluates to:

\[ \frac{5}{6}. \]

### Step 2: Evaluate the limit of the exponent as \( x \to 1 \)

Next, we need to evaluate the exponent:

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

Using the known limit:

\[ \lim_{t \to 0} \frac{1 - \cos(t)}{t^2} = \frac{1}{2}, \]

we can make a substitution \( t = x - 1 \). As \( x \to 1 \), \( t \to 0 \). Therefore, we have:

\[ \lim_{x \to 1} \frac{1 - \cos(x - 1)}{(x - 1)^2} = \lim_{t \to 0} \frac{1 - \cos(t)}{t^2} = \frac{1}{2}. \]

### Step 3: Combine the results

Now we can substitute back into the limit:

\[ \lim_{x \to 1} \left( \frac{5}{6} \right)^{\frac{1}{2}} = \sqrt{\frac{5}{6}}. \]

### Final Result

Thus, the final result is:

\[ \sqrt{\frac{5}{6}}. \]