Let `a = lim_(x->0) x cotx` and `b = lim_(x->0) xlog x,` then

Options

A. a =b
B. b > a
C. a =b+1
D. b =a +1

Answer

To solve the problem, we need to evaluate the limits for \( a \) and \( b \):

1. **Calculate \( a = \lim_{x \to 0} x \cot x \)**

We know that:
\[ \cot x = \frac{\cos x}{\sin x} \]
Therefore:
\[ x \cot x = x \cdot \frac{\cos x}{\sin x} = \frac{x \cos x}{\sin x} \]
As \( x \to 0 \), both the numerator and denominator approach 0, which gives us the indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule:

Differentiate the numerator and the denominator:
- Derivative of the numerator \( x \cos x \):
\[ \frac{d}{dx}(x \cos x) = \cos x - x \sin x \]
- Derivative of the denominator \( \sin x \):
\[ \frac{d}{dx}(\sin x) = \cos x \]

Now applying L'Hôpital's Rule:
\[ a = \lim_{x \to 0} \frac{\cos x - x \sin x}{\cos x} \]

Evaluating this limit as \( x \to 0 \):
\[ a = \frac{\cos(0) - 0 \cdot \sin(0)}{\cos(0)} = \frac{1 - 0}{1} = 1 \]

2. **Calculate \( b = \lim_{x \to 0} x \log x \)**

As \( x \to 0 \), \( \log x \to -\infty \), and thus \( x \log x \) approaches the indeterminate form \( 0 \cdot (-\infty) \). We can rewrite it as:
\[ b = \lim_{x \to 0} \frac{\log x}{\frac{1}{x}} \]
This is now in the form \( \frac{-\infty}{\infty} \), so we can apply L'Hôpital's Rule again:

Differentiate the numerator and the denominator:
- Derivative of \( \log x \):
\[ \frac{d}{dx}(\log x) = \frac{1}{x} \]
- Derivative of \( \frac{1}{x} \):
\[ \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \]

Now applying L'Hôpital's Rule:
\[ b = \lim_{x \to 0} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0} -x = 0 \]

3. **Conclusion:**

We have found:
\[ a = 1 \quad \text{and} \quad b = 0 \]
Now we can check the options:
- \( a = b \) → False (1 ≠ 0)
- \( b > a \) → False (0 < 1)
- \( a = b + 1 \) → True (1 = 0 + 1)
- \( b = a + 1 \) → False (0 ≠ 1 + 1)

Thus, the correct option is:
\[ a = b + 1 \]