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

Options

A. 1
B. `-1`
C. 2
D. 0

Answer

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

### Step 1: Rewrite the expression
We start by rewriting the expression using the definition of cotangent:
\[ \cot x = \frac{\cos x}{\sin x} \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \left( \frac{1}{x} - \frac{\cos x}{\sin x} \right) \]

### Step 2: Combine the fractions
Next, we find a common denominator to combine the fractions:
\[ \frac{1}{x} - \frac{\cos x}{\sin x} = \frac{\sin x - x \cos x}{x \sin x} \]
So now our limit becomes:
\[ \lim_{x \to 0} \frac{\sin x - x \cos x}{x \sin x} \]

### Step 3: Analyze the numerator
We need to analyze the numerator \( \sin x - x \cos x \). We can use Taylor series expansions for \( \sin x \) and \( \cos x \):
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]
\[ \cos x = 1 - \frac{x^2}{2} + O(x^4) \]
Substituting these into the expression for the numerator:
\[ \sin x - x \cos x = \left( x - \frac{x^3}{6} + O(x^5) \right) - x \left( 1 - \frac{x^2}{2} + O(x^4) \right) \]
This simplifies to:
\[ \sin x - x \cos x = x - \frac{x^3}{6} - x + \frac{x^3}{2} + O(x^5) = \left( \frac{1}{2} - \frac{1}{6} \right)x^3 + O(x^5) = \frac{1}{3}x^3 + O(x^5) \]

### Step 4: Substitute back into the limit
Now substituting back into our limit:
\[ \lim_{x \to 0} \frac{\frac{1}{3}x^3 + O(x^5)}{x \sin x} \]
We also need to expand \( \sin x \) in the denominator:
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]
So:
\[ x \sin x = x^2 - \frac{x^4}{6} + O(x^6) \]

### Step 5: Final limit calculation
Thus, our limit now looks like:
\[ \lim_{x \to 0} \frac{\frac{1}{3}x^3 + O(x^5)}{x^2 - \frac{x^4}{6} + O(x^6)} \]
As \( x \to 0 \), the leading term in the numerator is \( \frac{1}{3}x^3 \) and in the denominator is \( x^2 \). Therefore, we can simplify:
\[ \lim_{x \to 0} \frac{\frac{1}{3}x^3}{x^2} = \lim_{x \to 0} \frac{1}{3} x = 0 \]

### Conclusion
Thus, the value of the limit is:
\[ \boxed{0} \]