Solve X^m*(logX)^m,m,n are positive integers as limit X tends to zero?

To solve the limit of the expression \( X^m (\log X)^m \) as \( X \) approaches zero, we need to analyze how each component behaves in this scenario. Let's break it down step by step.

Understanding the Components

In the expression \( X^m \), as \( X \) approaches zero, \( X^m \) also approaches zero since \( m \) is a positive integer. On the other hand, \( \log X \) behaves differently. As \( X \) approaches zero from the positive side, \( \log X \) approaches negative infinity. Therefore, \( (\log X)^m \) will approach positive infinity because raising a negative number to an even power results in a positive value, and to an odd power results in a negative value.

Combining the Behaviors

Now, we need to consider the entire expression \( X^m (\log X)^m \). We have:

• As \( X \to 0^+ \), \( X^m \to 0 \).
• As \( X \to 0^+ \), \( (\log X)^m \to \infty \) (positive for even \( m \) and negative for odd \( m \)).

This creates a scenario where we have a product of something approaching zero and something approaching infinity. The key question is which of these two behaviors dominates the limit.

Applying L'Hôpital's Rule

To analyze this limit more rigorously, we can rewrite the expression in a form suitable for L'Hôpital's Rule. We can express \( X^m \) as \( e^{m \log X} \), leading to:

\[ X^m (\log X)^m = e^{m \log X} (\log X)^m \]

Now, we can rewrite the limit as:

\[ \lim_{X \to 0^+} \frac{(\log X)^m}{\frac{1}{X^m}} \]

This is now in the form of \( \frac{\infty}{\infty} \), which allows us to apply L'Hôpital's Rule. We differentiate the numerator and denominator:

Calculating Derivatives

The derivative of the numerator \( (\log X)^m \) using the chain rule is:

\[ m (\log X)^{m-1} \cdot \frac{1}{X} \]

The derivative of the denominator \( \frac{1}{X^m} \) is:

\[ -\frac{m}{X^{m+1}} \]

Applying L'Hôpital's Rule gives us:

\[ \lim_{X \to 0^+} \frac{m (\log X)^{m-1} \cdot \frac{1}{X}}{-\frac{m}{X^{m+1}}} \]

After simplifying, we find:

\[ \lim_{X \to 0^+} -X^m (\log X)^{m-1} \]

We can repeat this process, applying L'Hôpital's Rule \( m \) times, ultimately leading us to:

\[ \lim_{X \to 0^+} (-1)^m \cdot m! \cdot X^m \to 0 \]

Thus, regardless of whether \( m \) is even or odd, the limit approaches zero.

Final Result

In conclusion, the limit of \( X^m (\log X)^m \) as \( X \) approaches zero is:

\[ \lim_{X \to 0^+} X^m (\log X)^m = 0 \]

This result illustrates how the rapid decay of \( X^m \) outweighs the growth of \( (\log X)^m \) as \( X \) approaches zero. If you have any further questions or need clarification on any part of this process, feel free to ask!