v) \( \lim _{x \rightarrow 0}(1-\sin 2 x)^{\frac{3}{x}} \)

To evaluate the limit \( \lim _{x \rightarrow 0}(1-\sin 2 x)^{\frac{3}{x}} \), we can use the fact that as \( x \) approaches 0, \( \sin(2x) \) approaches \( 2x \). Therefore, we can write: \[ 1 - \sin(2x) \approx 1 - 2x \] Now we can express our limit as: \[ \lim_{x \rightarrow 0} (1 - \sin(2x))^{\frac{3}{x}} \approx \lim_{x \rightarrow 0} (1 - 2x)^{\frac{3}{x}} \] To simplify this, we will take the natural logarithm of the expression: \[ \ln L = \lim_{x \rightarrow 0} \frac{3}{x} \ln(1 - 2x) \] Using the approximation \( \ln(1 - u) \approx -u \) for small \( u \): \[ \ln(1 - 2x) \approx -2x \] Substituting this back into our limit gives: \[ \ln L \approx \lim_{x \rightarrow 0} \frac{3}{x} (-2x) = \lim_{x \rightarrow 0} -6 = -6 \] So, we have: \[ L = e^{-6} \] Thus, the limit evaluates to: \[ \lim _{x \rightarrow 0}(1-\sin 2 x)^{\frac{3}{x}} = e^{-6} \]