`Lt_(xto0)(x^(2)cosx)/(1-cosx)` is equal to

Options

A. 2
B. `(3)/(2)`
C. `-(3)/(2)`
D. 1

Answer

To solve the limit \( \lim_{x \to 0} \frac{x^2 \cos x}{1 - \cos x} \), we will follow these steps:

### Step 1: Identify the form of the limit
First, we substitute \( x = 0 \) into the expression:
\[ \frac{0^2 \cos(0)}{1 - \cos(0)} = \frac{0}{0} \]
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:
\[ \text{Numerator: } \frac{d}{dx}(x^2 \cos x) = 2x \cos x - x^2 \sin x \]
\[ \text{Denominator: } \frac{d}{dx}(1 - \cos x) = \sin x \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{2x \cos x - x^2 \sin x}{\sin x} \]

### Step 3: Substitute \( x = 0 \) again
Now, substituting \( x = 0 \) again gives:
\[ \frac{2(0) \cos(0) - 0^2 \sin(0)}{\sin(0)} = \frac{0}{0} \]
We still have an indeterminate form, so we apply L'Hôpital's Rule again.

### Step 4: Differentiate again
Differentiate the numerator and denominator again:
\[ \text{Numerator: } \frac{d}{dx}(2x \cos x - x^2 \sin x) = 2\cos x - 2x \sin x - (2x \sin x + x^2 \cos x) = 2\cos x - 4x \sin x - x^2 \cos x \]
\[ \text{Denominator: } \frac{d}{dx}(\sin x) = \cos x \]
Thus, we have:
\[ \lim_{x \to 0} \frac{2\cos x - 4x \sin x - x^2 \cos x}{\cos x} \]

### Step 5: Substitute \( x = 0 \) again
Now substituting \( x = 0 \):
\[ \frac{2\cos(0) - 4(0)\sin(0) - 0^2 \cos(0)}{\cos(0)} = \frac{2 \cdot 1 - 0 - 0}{1} = 2 \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to 0} \frac{x^2 \cos x}{1 - \cos x} = 2 \]