We have $f(x)$:
$\lim_{x\to\infty} (f^2(x)) = L > 0 $
We need to prove/contradict that:
$\lim_{x\to\infty} f(x) = \sqrt{L} > 0$
I didn't find any counterexample and I'm having trouble proving it...any help?
2026-03-28 14:36:58.1774708618
Prove that a limit exist for a function
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This is a counterexample:
$$f(x)=(-1)^{\lfloor x\rfloor}$$