91Ó°ÊÓ

A limit in calculus helps us understand how a function behaves as the input approaches a certain point. Let's consider the function \( f(x) = x(\ln x)^2 \) when \( x > 0 \) and 0 for \( x = 0 \).
In this problem, we need to find the limit of \( f(x) \) as \( x \) approaches 0 from the right, denoted as \( \lim_{x \to 0^+} \).
This means we are interested in how \( f(x) \) behaves as \( x \) gets closer and closer to 0 from positive values.

• As \( x \to 0^+ \), \( x \) becomes very small positive values.
• We substitute \( y = \ln x \) to simplify our analysis, so \( x = e^y \).
• Thus, the expression becomes \( e^y y^2 \) as \( y \to -\infty \).

The goal is to analyze \( \lim_{y \to -\infty} (e^y y^2) \). Here, \( e^y \) tends to 0 very quickly, overpowering the term \( y^2 \) that tries to grow. As a result, the entire product approaches 0. This shows the solution of the limit is \( 0 \).

The natural logarithm, denoted as \( \ln x \), is a special logarithm with base \( e \) (a mathematical constant approximately equal to 2.71828).
In this function \( f(x) = x(\ln x)^2 \), the natural logarithm plays a critical role.

• As \( x \to 0^+ \), \( \ln x \to -\infty \) because \( \, \ln \, 0 \) is undefined and small positive \( x \) leads \( \ln x \) towards negative infinity.
• The square \( (\ln x)^2 \) changes the sign, leading the output towards positive infinity.

Yet, despite this significance, the product \( x(\ln x)^2 \) still trends to 0 due to \( x \) being very small, demonstrating the power of exponential decay in terms like \( e^y \) when \( y \) is the logarithm at the range we are interested in. This behavior of \( \ln x \) is crucial in understanding the continuity behavior of the function.

Continuity at a point, specifically from the right in this scenario, implies smoothness and a lack of sudden jumps or breaks in the graph of a function.
To declare a function continuous from the right at a point such as \( x = 0 \), we need:

• \( \lim_{x \to 0^+} f(x) = f(0) \)

This essentially means that as we approach 0 from values greater than 0, the limit of the function should be the same as its actual value at 0.
In our case, \( \lim_{x \to 0^+} x(\ln x)^2 = 0 \), which matches \( f(0) = 0 \).
Thus, there is no disruption or jump in the value of the function as \( x \) smoothly approaches 0 from the right. This guarantees the function \( f(x) \) is right continuous at \( x = 0 \). Such analysis not only confirms the continuity but also strengthens our understanding of the behavior of functions as they approach certain critical values from one direction.