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susus · Jan 31, 2011, 08:55 AM

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

ebaines · Jan 31, 2011, 09:42 AM

Suggestion: if

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

Then

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

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

galactus · Jan 31, 2011, 09:46 AM

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 $e^{\frac{-1}{2}}=\frac{1}{\sqrt{e}}$