The value of `lim_(x to 1){(x^(3) + 2x^(2) + x+1)/(x^(2) + 2x + 3)}^((1- cos(x-1))/(x-1)^(2))` is equal to:

Options

A. e
B. `e^(1//2)`
C. 1
D. `sqrt(5//6)`

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 break it down into steps.

### Step 1: Evaluate the limit of the exponent

First, we need to evaluate the limit

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

Using the Taylor series expansion for \(\cos(x)\) around \(x = 0\):

\[ \cos(x) \approx 1 - \frac{x^2}{2} + O(x^4), \]

we can substitute \(x - 1\) for \(x\):

\[ 1 - \cos(x - 1) \approx \frac{(x - 1)^2}{2} + O((x - 1)^4). \]

Thus,

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

### Step 2: Evaluate the base of the limit

Next, we evaluate the limit of the base:

\[ \lim_{x \to 1} \frac{x^3 + 2x^2 + x + 1}{x^2 + 2x + 3}. \]

Substituting \(x = 1\):

\[ \frac{1^3 + 2 \cdot 1^2 + 1 + 1}{1^2 + 2 \cdot 1 + 3} = \frac{1 + 2 + 1 + 1}{1 + 2 + 3} = \frac{5}{6}. \]

### Step 3: Combine the results

Now we can combine the results from Steps 1 and 2 into the original limit:

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

### Final Answer

Thus, the value of the limit is

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