We know that $\lim_{n\to\infty}(-1)^n=(-1)^{\infty}$ doesn't exist. Now take $\lim_{n\to\infty}(-1)^{2n}=(-1)^{\infty}$. This limit exists, because $\lim_{n\to\infty}(-1)^{2n}=\lim_{n\to\infty}((-1)^2)^n=1$. Does this mean that $(-1)^{\infty}$ is an indeterminate form?
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