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lim as x goes to inf (cosx)^(1/x^2)
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Recall that cosine is bounded by -1 and 1. \[\large\rm -1\le \cos x\le 1\]If we raise each side to the 1/x^2 power, we get,\[\large\rm (-1)^{1/x^2}\le (\cos x)^{1/x^2}\le 1^{1/x^2}\] \(\large\rm (-1)^{1/x^2}\to1\) \(\large\rm 1^{1/x^2}\to1\) Squeeze Theorem does the rest, ya? :)
no
the limit does not exist
\[\lim_{x\to\infty}(\cos x)^{1/x^2}=\exp\left(\lim_{x\to\infty}\ln\frac{\cos x}{x^2}\right)=\exp\left(\color{red}{\lim_{x\to\infty}\ln\cos x}-\lim_{x\to\infty}x^2\right)\]
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