We are tasked with evaluating the following limit:

$\lim_{x \to 0^+} x \ln(4x^2)$

Step 1: Simplifying the Expression

We can start by simplifying the logarithmic term:

$\ln(4x^2) = \ln(4) + \ln(x^2) = \ln(4) + 2\ln(x)$

So, the limit becomes:

$\lim_{x \to 0^+} x \left( \ln(4) + 2\ln(x) \right)$

Distribute $x$:

$\lim_{x \to 0^+} \left( x \ln(4) + 2x \ln(x) \right)$

Step 2: Evaluating Each Term

1. Term 1: $x \ln(4)$

Since $\ln(4)$ is a constant, this term behaves as:

$\lim_{x \to 0^+} x \ln(4) = 0$

2. Term 2: $2x \ln(x)$

This term requires more careful handling. As $x \to 0^+$, we need to determine the behavior of $x \ln(x)$. Notice that $\ln(x)$ becomes very negative as $x$ approaches 0 from the right. Let's evaluate this term using L'Hopital's Rule, since it has the indeterminate form $0 \times (-\infty)$.

Rewriting it as:

$\lim_{x \to 0^+} 2x \ln(x) = \lim_{x \to 0^+} \frac{2 \ln(x)}{1/x}$

Now, apply L'Hopital's Rule (which is applicable for indeterminate forms of type $\frac{-\infty}{\infty}$):

Differentiate the numerator and denominator:

• Derivative of $2 \ln(x)$ is $\frac{2}{x}$
• Derivative of $1/x$ is $-\frac{1}{x^2}$

Applying L'Hopital's Rule:

$\lim_{x \to 0^+} \frac{2/x}{-1/x^2} = \lim_{x \to 0^+} -2x = 0$

Thus, the second term $\lim_{x \to 0^+} 2x \ln(x) = 0$.

Step 3: Conclusion

Both terms tend to 0, so:

$\lim_{x \to 0^+} x \ln(4x^2) = 0$

Final Answer:

$\boxed{0}$

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Would you like more details on L'Hopital's Rule or the behavior of logarithmic functions near zero? Here are some related questions:

1. How does L'Hopital's Rule work with limits of the form $0 \times \infty$?
2. What is the behavior of $x \ln(x)$ as $x \to 0^+$?
3. What happens when we have limits involving other logarithmic functions, like $\ln(x^n)$?
4. How does logarithmic growth compare to polynomial or exponential growth as $x$ approaches 0 or infinity?
5. Can we apply L'Hopital's Rule to limits involving trigonometric functions?

Tip: When encountering logarithmic limits like $x \ln(x)$, it's often useful to rewrite the expression in a form suitable for L'Hopital's Rule.