How do you find the limit of $ln(lnt)$ as $t->oo$ ?

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Answer 1 · 2017-01-02T17:28:22

The answer is $infty$

Since $ln(t)->infty$ as $t->infty$ , it follows that $ln(ln(t))->infty$ as $t->infty$ (though very, very, slowly).

To illustrate how slowly $ln(ln(t))->infty$ as $t->infty$ , you might ask: how big should $t$ be so that $ln(ln(t))>10$ ? (for example)

To answer this, solve the inequality $ln(ln(t))>10$ by exponentiation: $ln(ln(t))>10 <=> ln(t)>e^(10) <=> t>e^(e^(10)) approx 9.4 times 10^(9565)$

In general, in order for $ln(ln(t))$ to be bigger than an arbitrary $M>0$ , you need to choose $t$ so large that $t>e^(e^(M))$ .

How do you find the limit of ln(lnt)ln(lnt) as t->oot->oo?