${\lim }_{x\to 0}(\cos x{)}^{1/x^2}$

Solution

1. Let $L={\lim }_{x\to 0}(\cos x{)}^{1/x^2}$, which is an indeterminate form of type $1^{\infty }$.
2. Taking the natural logarithm, we have $\ln L={\lim }_{x\to 0}\frac{\ln (\cos x)}{x^2}$.
3. Using the Taylor series expansion $\cos x\approx 1-\frac{x^2}{2}$, the expression becomes ${\lim }_{x\to 0}\frac{\ln (1-x^2/2)}{x^2}$.
4. Applying the approximation $\ln (1+u)\approx u$ for small $u$, we get ${\lim }_{x\to 0}\frac{-x^2/2}{x^2}=-1/2$.
5. Since $\ln L=-1/2$, it follows that $L=e^{-1/2}$.
6. Hence the answer is (A).