The value of `lim_(xrarr0^(+))((x cot x)+(x lnx))` is equal to

Options

A. 1
B. 2
C. 3
D. 0

Answer

To find the limit \( \lim_{x \to 0^+} (x \cot x + x \ln x) \), we can break it down into two parts: \( x \cot x \) and \( x \ln x \).

### Step 1: Analyze \( x \cot x \)

We know that:
\[ \cot x = \frac{\cos x}{\sin x} \]
As \( x \to 0^+ \), \( \cot x \) approaches infinity since \( \sin x \) approaches 0. Therefore, we have:
\[ x \cot x = x \cdot \frac{\cos x}{\sin x} \]
This results in the form \( 0 \cdot \infty \). To resolve this, we rewrite it:
\[ x \cot x = \frac{x \cos x}{\sin x} \]
Now, as \( x \to 0^+ \), both the numerator and denominator approach 0, resulting in the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule to \( x \cot x \)

Using L'Hôpital's Rule, we differentiate the numerator and denominator:
- Derivative of the numerator \( x \cos x \) is \( \cos x - x \sin x \).
- Derivative of the denominator \( \sin x \) is \( \cos x \).

Thus, we have:
\[ \lim_{x \to 0^+} \frac{x \cos x}{\sin x} = \lim_{x \to 0^+} \frac{\cos x - x \sin x}{\cos x} \]

Now substituting \( x = 0 \):
\[ = \frac{1 - 0}{1} = 1 \]

### Step 3: Analyze \( x \ln x \)

Next, we consider \( x \ln x \). As \( x \to 0^+ \), \( \ln x \) approaches \( -\infty \). Therefore, \( x \ln x \) also results in the form \( 0 \cdot (-\infty) \). We rewrite it:
\[ x \ln x = \frac{\ln x}{1/x} \]
This gives us the form \( \frac{-\infty}{\infty} \).

### Step 4: Apply L'Hôpital's Rule to \( x \ln x \)

Using L'Hôpital's Rule again:
- Derivative of the numerator \( \ln x \) is \( \frac{1}{x} \).
- Derivative of the denominator \( \frac{1}{x} \) is \( -\frac{1}{x^2} \).

Thus, we have:
\[ \lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} -x = 0 \]

### Step 5: Combine the Results

Now, we combine the results from both parts:
\[ \lim_{x \to 0^+} (x \cot x + x \ln x) = 1 + 0 = 1 \]

### Final Answer

Thus, the value of \( \lim_{x \to 0^+} (x \cot x + x \ln x) \) is:
\[ \boxed{1} \]