`lim_(xrarr0^(+))(x^(n)lnx), n gt0`

Options

A. does not exist
B. exists and is zero
C. exists and is 1
D. exists and is `e^(-1)`

Answer

To solve the limit \( \lim_{x \to 0^+} (x^n \ln x) \) where \( n > 0 \), we can follow these steps:

### Step 1: Substitute \( x \)
We can substitute \( x \) with \( e^{-t} \), where \( t \to \infty \) as \( x \to 0^+ \). This gives us:
\[ \lim_{x \to 0^+} (x^n \ln x) = \lim_{t \to \infty} ((e^{-t})^n \ln(e^{-t})) \]

### Step 2: Simplify the expression
The expression simplifies to:
\[ \ln(e^{-t}) = -t \]
Thus, we have:
\[ \lim_{t \to \infty} (e^{-nt} \cdot (-t)) = -\lim_{t \to \infty} \frac{t}{e^{nt}} \]

### Step 3: Analyze the limit
Now we need to evaluate:
\[ -\lim_{t \to \infty} \frac{t}{e^{nt}} \]
This is an indeterminate form of type \( \frac{\infty}{\infty} \), so we can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
Differentiating the numerator and denominator:
- The derivative of the numerator \( t \) is \( 1 \).
- The derivative of the denominator \( e^{nt} \) is \( n e^{nt} \).

Thus, we have:
\[ -\lim_{t \to \infty} \frac{1}{n e^{nt}} = -\frac{1}{n} \cdot \lim_{t \to \infty} \frac{1}{e^{nt}} \]

### Step 5: Evaluate the limit
As \( t \to \infty \), \( e^{nt} \to \infty \), so:
\[ \lim_{t \to \infty} \frac{1}{e^{nt}} = 0 \]
Thus, we conclude:
\[ -\frac{1}{n} \cdot 0 = 0 \]

### Final Result
Therefore, the limit is:
\[ \lim_{x \to 0^+} (x^n \ln x) = 0 \]

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