\( \lim _ { x \rightarrow 0 } \sqrt { 5 x } \ln ( \sin ( x ) ) \)

To evaluate \( \lim _ { x \rightarrow 0 } \sqrt { 5 x } \ln ( \sin ( x ) ) \), let's first consider the behavior of each part of the expression as \( x \) approaches \( 0 \). As \( x \) approaches \( 0 \): - \( \sqrt{5x} \) approaches \( 0 \). - \( \sin(x) \) approaches \( x \), so \( \ln(\sin(x)) \approx \ln(x) \) as \( x \) approaches \( 0 \). Now we can rewrite the limit using the approximation for \( \ln(\sin(x)) \): \[ \lim _ { x \rightarrow 0 } \sqrt { 5 x } \ln ( \sin ( x ) ) \approx \lim _ { x \rightarrow 0 } \sqrt { 5 x } \ln(x). \] Next, we rewrite \( \ln(x) \) in a different form to consider the limit: \[ \ln(x) = \ln(\sqrt{x^2}) = \frac{1}{2} \ln(x^2) = 2 \ln(x). \] However, let's focus on the limit directly: \[ \lim_{x \to 0} \sqrt{5x} \ln(x) = \lim_{x \to 0} \sqrt{5} \cdot \sqrt{x} \cdot \ln(x). \] To analyze \( \sqrt{x} \ln(x) \), notice that as \( x \) approaches \( 0^+ \), \( \ln(x) \) approaches \( -\infty \) while \( \sqrt{x} \) approaches \( 0 \). Therefore, we are left with an indeterminate form of type \( 0 \cdot (-\infty) \). To resolve this, we can rewrite it as: \[ \sqrt{x} \ln(x) = \frac{\ln(x)}{1/\sqrt{x}}. \] Now, we will evaluate this limit using L'Hôpital's Rule since it's of the \( \frac{-\infty}{\infty} \) form: \[ \lim_{x \to 0^+} \frac{\ln(x)}{1/\sqrt{x}}. \] Taking the derivative of the numerator and the denominator: - The derivative of \( \ln(x) \) is \( \frac{1}{x} \). - The derivative of \( \frac{1}{\sqrt{x}} \) is \( -\frac{1}{2} x^{-3/2} \). Applying L'Hôpital's Rule results in: \[ \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{2} x^{-3/2}} = \lim_{x \to 0^+} -2\sqrt{x} = 0. \] Therefore, we conclude: \[ \lim _ { x \rightarrow 0 } \sqrt { 5 x } \ln ( \sin ( x ) ) = 0. \]