Using L' Hospital's rule, evaluate:
`lim_(x to 0) ((1)/(x^(2))-(cotx)/(x))`

Answer

To evaluate the limit using L'Hôpital's rule, we start with the expression:

\[ \lim_{x \to 0} \left( \frac{1}{x^2} - \frac{\cot x}{x} \right) \]

### Step 1: Combine the fractions

We can combine the two fractions into a single fraction:

\[ \lim_{x \to 0} \left( \frac{1}{x^2} - \frac{\cot x}{x} \right) = \lim_{x \to 0} \left( \frac{x - x^2 \cot x}{x^2} \right) \]

### Step 2: Rewrite cotangent

Recall that \(\cot x = \frac{\cos x}{\sin x}\), thus we can rewrite the limit as:

\[ \lim_{x \to 0} \left( \frac{x - x^2 \frac{\cos x}{\sin x}}{x^2} \right) = \lim_{x \to 0} \left( \frac{x \sin x - x^2 \cos x}{x^2 \sin x} \right) \]

### Step 3: Simplify the expression

Now, we can simplify the numerator:

\[ \lim_{x \to 0} \left( \frac{x \sin x - x^2 \cos x}{x^2 \sin x} \right) \]

As \(x\) approaches \(0\), both the numerator and denominator approach \(0\), which 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. We differentiate the numerator and the denominator separately:

1. Differentiate the numerator:
\[ \frac{d}{dx}(x \sin x - x^2 \cos x) = \sin x + x \cos x - (2x \cos x - x^2 \sin x) = \sin x + x \cos x - 2x \cos x + x^2 \sin x \]
Simplifying gives:
\[ \sin x - x \cos x + x^2 \sin x \]

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

Now we have:

\[ \lim_{x \to 0} \frac{\sin x - x \cos x + x^2 \sin x}{2x \sin x + x^2 \cos x} \]

### Step 5: Evaluate the limit again

Substituting \(x = 0\):

- The numerator becomes:
\[ \sin(0) - 0 \cdot \cos(0) + 0^2 \cdot \sin(0) = 0 \]
- The denominator becomes:
\[ 2 \cdot 0 \cdot \sin(0) + 0^2 \cdot \cos(0) = 0 \]

This is still an indeterminate form, so we apply L'Hôpital's Rule again.

### Step 6: Differentiate again

1. Differentiate the new numerator:
\[ \frac{d}{dx}(\sin x - x \cos x + x^2 \sin x) = \cos x - (\cos x - x \sin x) + (2x \sin x + x^2 \cos x) \]
Simplifying gives:
\[ 2x \sin x + x^2 \cos x \]

2. Differentiate the new denominator:
\[ \frac{d}{dx}(2x \sin x + x^2 \cos x) = 2 \sin x + 2x \cos x + (2x \cos x - x^2 \sin x) \]
Simplifying gives:
\[ 4x \cos x + 2 \sin x - x^2 \sin x \]

### Step 7: Evaluate the limit again

Substituting \(x = 0\):

- The numerator becomes:
\[ 2 \cdot 0 \cdot \sin(0) + 0^2 \cdot \cos(0) = 0 \]
- The denominator becomes:
\[ 4 \cdot 0 \cdot \cos(0) + 2 \cdot \sin(0) - 0^2 \cdot \sin(0) = 0 \]

### Step 8: Final differentiation

Apply L'Hôpital's Rule one last time:

1. Differentiate the numerator:
\[ \frac{d}{dx}(2x \sin x) = 2 \sin x + 2x \cos x \]

2. Differentiate the denominator:
\[ \frac{d}{dx}(4x \cos x + 2 \sin x) = 4 \cos x - 4x \sin x + 2 \cos x = 6 \cos x - 4x \sin x \]

### Step 9: Evaluate the limit

Now we can evaluate:

\[ \lim_{x \to 0} \frac{2 \sin x + 2x \cos x}{6 \cos x - 4x \sin x} \]

Substituting \(x = 0\):

- The numerator becomes:
\[ 2 \cdot \sin(0) + 2 \cdot 0 \cdot \cos(0) = 0 \]
- The denominator becomes:
\[ 6 \cdot \cos(0) - 4 \cdot 0 \cdot \sin(0) = 6 \]

Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{0}{6} = 0 \]

### Final Answer

The limit is:

\[ \boxed{0} \]