\( \Rightarrow \lim _ { x \rightarrow 0 ^ { + } } 4 x \ln x = \)

To evaluate the limit \( \lim _ { x \rightarrow 0 ^ { + } } 4 x \ln x \), we can make a substitution. As \( x \) approaches \( 0^+ \), \( \ln x \) approaches \( -\infty \), but at a slower rate than \( x \) approaches \( 0 \). Thus, we can rewrite \( 4x \ln x = 4 \frac{\ln x}{\frac{1}{x}} \). As \( x \) approaches \( 0^+ \), \( \ln x \) approaches \( -\infty \) and \( \frac{1}{x} \) approaches \( +\infty \). This is an indeterminate form of type \( \frac{-\infty}{+\infty} \), and we can apply L'Hôpital's Rule: Taking the derivative of the numerator and the denominator, we have: - Derivative of \( \ln x \): \( \frac{1}{x} \) - Derivative of \( \frac{1}{x} \): \( -\frac{1}{x^2} \) Now applying L'Hôpital's Rule: \[ \lim_{x \rightarrow 0^+} 4 x \ln x = 4 \lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}} = 4 \lim_{x \rightarrow 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = 4 \lim_{x \rightarrow 0^+} -x = 0. \] Therefore, the limit is: \[ \lim _ { x \rightarrow 0 ^ { + } } 4 x \ln x = 0. \]