If I consider that $\lim_{x\to\infty}\frac{\log{x}}{x^a}=0$ $\forall a>0$, then can I write that $\exists M>0$ such that $\frac{\log{x}}{x^a}<1$ $\forall x>M$? Is it the definition of limit in which I use $\epsilon=1$, right?
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Yes, but that $M$ will depend upon the actual value of $a$. The $M_1$ required that $\frac {\log x}{x^1} < 1$ when $x > M_1$ will be a different value then then $M_{0.001}$ required that $\frac {\log x}{x^{0.001}} < 1$ when $x > M_{0.0001}$. – fleablood May 08 '21 at 15:28
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Possible duplicate: Compute $\lim\limits_{x \to \infty} \frac{\log(x)}{x^a}$ – robjohn Dec 04 '21 at 06:46
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Yes ; by the epsilon delta definition of a limit since you know $$\lim_{x\to \infty} \frac{\log(x)}{x^{\alpha}}=0$$ for some alpha $>0$ then by definition it means that $\forall \epsilon>0$ there exists a $M$ such that for all $x>M$ we have $|\log(x)/x^{\alpha}| < \epsilon$ now you can just let epsilon be $1$ . Note that the $M$ will not be the same for all alpha.
Vivaan Daga
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Ok thanks but if I want to traslate the limit only in an inequality I can say that eventually $\frac{\log{x}}{x^a}<1$, right? What's change is only the value starting from which the inequality holds? – Sisi May 08 '21 at 15:33
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Yes you can say that the equality holds eventually for large enough $x$ . – Vivaan Daga May 08 '21 at 15:35