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I saw the following definition on youtube.

Let $f$ be a function defined on some interval $(a, +\infty)$

$\lim\limits_{x \to +\infty} f(x)= +\infty$ (1)

means $\forall N > 0$ $\exists M > 0$ such that $x > M$ implies $f(x) > N$.

Is this a correct and general definition for equation (1)? For example, the function $f$ defined below doesn't satisfy the above definition, but it is obviously divergent, as $x \to +\infty$.

$f(x) \equiv \log(x) \sin^2(x)$, $x \in (0, +\infty)$

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    $\begingroup$ Yes, that is the definition. The example $f$ that you gave doesn't have a limit that is $+\infty$. $\endgroup$
    – NDB
    Commented Jul 18, 2023 at 3:21
  • $\begingroup$ Thanks! Let me try to understand this. So the divergence of a function doesn't necessarily mean it asymptotically approaches the infinity, as $x \to +\infty$, correct? $\endgroup$
    – Jerry
    Commented Jul 18, 2023 at 3:33
  • $\begingroup$ Or, in another word, the equation (1) is not the definition of the divergence of a function, as $x \to +\infty$. $\endgroup$
    – Jerry
    Commented Jul 18, 2023 at 3:35
  • $\begingroup$ That’s correct. Although if someone writes not just “is divergent”, but “diverges to positive infinity”, they would mean equation (1). $\endgroup$
    – Michael Weiss
    Commented Jul 18, 2023 at 3:36
  • $\begingroup$ Yes, a limit that is $+\infty$ is a special case of those called "divergent" in calculus. It is closer in properties to limits that exist, that are called "convergent" than to other divergent limits. $\endgroup$
    – NDB
    Commented Jul 18, 2023 at 3:37

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