lim (cosx)^(1/x^2)
x-->0

Suggestion: if


L = \displaystyle \lim_{x \to 0} (\cos x)^{1/n^2}


Then


\ln(L} = \displaystyle \lim_{x \to 0} \ln (\cos x)^{1/n^2} = \displaystyle \lim_{x \to 0} \frac { \ln (\cos x)}{n^2}


Now use de Moivre's theorem to find the limit of ln(L). Then L is e raised to that value.

A trick we can use is by observing that cos(x)\sim 1-\frac{x^{2}}{2} for x\to 0

\lim_{x\to 0}\left(1-\frac{x^{2}}{2}\right)^{\frac{1}{x^{2}}}

Now, make the sub:

t=\frac{1}{x^{2}} and we get:

\lim_{t\to \infty}\left(1-\frac{1}{2t}\right)^{t}

Notice, this looks mighty similar to the famous 'e' limit:

e=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{t}

except we have a \frac{-1}{2t}

Thus, the limit is e^{\frac{-1}{2}}=\frac{1}{\sqrt{e}}