The value of `lim_(x to 0) ((1)/(x^(2)) - cot x)` equals

Options

A. 1
B. 0
C. `oo`
D. Does not exist

Answer

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

### Step 1: Rewrite the limit
We start with the expression:
\[ \lim_{x \to 0} \left( \frac{1}{x^2} - \cot x \right) \]
Recall that \( \cot x = \frac{\cos x}{\sin x} \). Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \left( \frac{1}{x^2} - \frac{\cos x}{\sin x} \right) \]

### Step 2: Find a common denominator
Next, we find a common denominator for the two fractions:
\[ \lim_{x \to 0} \left( \frac{\sin x - x^2 \cos x}{x^2 \sin x} \right) \]

### Step 3: Evaluate the limit
Now, we substitute \( x = 0 \):
\[ \sin(0) - 0^2 \cos(0) = 0 - 0 = 0 \]
and the denominator becomes:
\[ 0^2 \sin(0) = 0 \cdot 0 = 0 \]
This gives us the indeterminate form \( \frac{0}{0} \).

### Step 4: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator:
\[ \text{Numerator: } \frac{d}{dx}(\sin x - x^2 \cos x) = \cos x - (2x \cos x - x^2 \sin x) \]
This simplifies to:
\[ \cos x - 2x \cos x + x^2 \sin x \]
\[ = \cos x (1 - 2x) + x^2 \sin x \]

For the denominator:
\[ \text{Denominator: } \frac{d}{dx}(x^2 \sin x) = 2x \sin x + x^2 \cos x \]

### Step 5: Substitute \( x = 0 \) again
Now we evaluate the limit again:
\[ \lim_{x \to 0} \frac{\cos x (1 - 2x) + x^2 \sin x}{2x \sin x + x^2 \cos x} \]
Substituting \( x = 0 \):
\[ \frac{1(1 - 0) + 0}{0 + 0} = \frac{1}{0} \]
This indicates that the limit approaches infinity.

### Conclusion
Thus, we conclude that:
\[ \lim_{x \to 0} \left( \frac{1}{x^2} - \cot x \right) = \infty \]