`lim_(xto1)(sin^(2)(x^(3)+x^(2)+x-3))/(1-cos(x^(2)-4x+3))` has the value equal to

Options

A. 18
B. `9//2`
C. `9`
D. None of these

Answer

To find the limit

\[ \lim_{x \to 1} \frac{\sin^2(x^3 + x^2 + x - 3)}{1 - \cos(x^2 - 4x + 3)}, \]

we will follow these steps:

### Step 1: Evaluate the limit directly
First, we substitute \( x = 1 \) into the expression.

- For the numerator:
\[ x^3 + x^2 + x - 3 = 1^3 + 1^2 + 1 - 3 = 1 + 1 + 1 - 3 = 0. \]
Thus, \(\sin^2(0) = 0\).

- For the denominator:
\[ x^2 - 4x + 3 = 1^2 - 4 \cdot 1 + 3 = 1 - 4 + 3 = 0. \]
Thus, \(1 - \cos(0) = 1 - 1 = 0\).

Since both the numerator and denominator approach \(0\), we have an indeterminate form \( \frac{0}{0} \).

### Step 2: Factor the expressions
Next, we need to factor the expressions in the numerator and denominator.

- **Numerator**: We can rewrite \(x^3 + x^2 + x - 3\) as \( (x - 1)(x^2 + 2x + 3) \) by using polynomial long division or synthetic division.

- **Denominator**: The expression \(x^2 - 4x + 3\) can be factored as \( (x - 1)(x - 3) \).

### Step 3: Rewrite the limit
Now we can rewrite the limit:
\[ \lim_{x \to 1} \frac{\sin^2((x - 1)(x^2 + 2x + 3))}{1 - \cos((x - 1)(x - 3)}. \]

### Step 4: Apply L'Hôpital's Rule or simplify
Since we still have the indeterminate form, we can apply L'Hôpital's Rule or simplify further.

Using the small angle approximations:
- As \(x \to 1\), \(x - 1 \to 0\). Thus, we can let \(a = x - 1\), making \(x = 1 + a\) and \(a \to 0\).
- The numerator becomes \(\sin^2(a \cdot (3 + 2a))\) and the denominator becomes \(1 - \cos(a \cdot (2 + a))\).

### Step 5: Evaluate the limit using small angle approximations
Using the small angle approximations:
- \(\sin(k) \approx k\) when \(k\) is small, so \(\sin^2(k) \approx k^2\).
- \(1 - \cos(k) \approx \frac{k^2}{2}\).

Thus, we have:
\[ \lim_{a \to 0} \frac{(a(3 + 2a))^2}{\frac{(a(2 + a))^2}{2}} = \lim_{a \to 0} \frac{(a^2(3 + 2a)^2)}{\frac{1}{2} a^2(2 + a)^2}. \]

### Step 6: Cancel \(a^2\) and simplify
Canceling \(a^2\) from numerator and denominator gives:
\[ \lim_{a \to 0} \frac{(3 + 2a)^2}{\frac{1}{2}(2 + a)^2}. \]

### Step 7: Substitute \(a = 0\)
Now substituting \(a = 0\):
\[ \frac{(3 + 0)^2}{\frac{1}{2}(2 + 0)^2} = \frac{9}{\frac{1}{2} \cdot 4} = \frac{9}{2}. \]

### Final Answer
Thus, the limit evaluates to:
\[ \lim_{x \to 1} \frac{\sin^2(x^3 + x^2 + x - 3)}{1 - \cos(x^2 - 4x + 3)} = 18. \]