The value of `lim_(xto0)(1/(x^(2))-cotx)` is

Options

A. `0`
B. `1`
C. `1/4`
D. None of these

Answer

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

### Step 1: Rewrite Cotangent
We start by rewriting the cotangent function in terms of sine and cosine:
\[ \cot x = \frac{\cos x}{\sin x} \]
Thus, we can express 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
Substituting \( x = 0 \) directly into the expression gives us:
\[ \frac{0 - 0}{0} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
We differentiate the numerator and the denominator:
- The derivative of the numerator \( \sin x - x^2 \cos x \) is:
\[ \cos x - (2x \cos x - x^2 \sin x) = \cos x - 2x \cos x + x^2 \sin x \]
- The derivative of the denominator \( x^2 \sin x \) is:
\[ 2x \sin x + x^2 \cos x \]

### Step 5: Rewrite the Limit
Now we rewrite the limit using the derivatives:
\[ \lim_{x \to 0} \frac{\cos x - 2x \cos x + x^2 \sin x}{2x \sin x + x^2 \cos x} \]

### Step 6: Evaluate the New Limit
Substituting \( x = 0 \) again gives:
\[ \frac{1 - 0 + 0}{0 + 0} = \frac{1}{0} \]
This indicates that the limit approaches infinity.

### Conclusion
Thus, the limit \( \lim_{x \to 0} \left( \frac{1}{x^2} - \cot x \right) \) is undefined (or approaches infinity).