Answer is: option1

\( e^{-2} \)

Solution:

Define:

\[ L = \lim_{y \to 0^+} \left[\cos(2y)\right]^{\frac{1}{y^2}} \]

Taking the natural logarithm:

\[ \ln L = \lim_{y \to 0^+} \frac{\ln(\cos(2y))}{y^2} \]

Since \( \ln(\cos(2y)) \to \ln(1) = 0 \) and \( y^2 \to 0 \), we get an indeterminate form \( \frac{0}{0} \). Applying L'Hôpital's Rule by differentiating the numerator and denominator:

Derivative of \( \ln(\cos(2y)) \):

\[ \frac{d}{dy} \ln(\cos(2y)) = \frac{-2\sin(2y)}{\cos(2y)} = -2\tan(2y) \]

Derivative of \( y^2 \):

\[ \frac{d}{dy} y^2 = 2y \]

Now applying L'Hopital's Rule:

\[ \ln L = \lim_{y \to 0^+} \frac{-2\tan(2y)}{2y} \]

Simplifying:

\[ \ln L = \lim_{y \to 0^+} \frac{-\tan(2y)}{y} \]

Using the small-angle approximation \( \tan(2y) \approx 2y \) as \( y \to 0 \):

\[ \ln L = \lim_{y \to 0^+} \frac{-2y}{y} = -2 \]

Since \( L = e^{\ln L} \), we get:

\[ L = e^{-2} \]

Thus, the given limit evaluates to: \( e^{-2} \)