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

Answer 1

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))#.

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Answer 2

Christopher Allers

To find the limit of ln(lnt) as t approaches infinity, we can use the properties of logarithms and limits.

Let's substitute u = lnt, so t = e^u.

As t approaches infinity, u also approaches infinity.

Now, we can rewrite the expression as ln(u).

The limit of ln(u) as u approaches infinity is infinity.

Therefore, the limit of ln(lnt) as t approaches infinity is also infinity.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit of #ln(lnt)# as #t-&gt;oo#?

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