Find this limit: lim arctan(n!) , n->infinity ?

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I don't understand how to find it and if exist
thanks

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Answer 1 · 2018-05-18T22:40:47

$lim_(n->oo) arctan(n!) = pi/2$

As $n$ becomes arbitrarily large, $n!$ also becomes very big. In other words, it approaches infinity.

$n->oo => n! ->oo$

Of course, $n!$ is infinitely bigger than $n$ , but for what we care, it's the same.

So we have

$lim_(n->oo) arctan(n!) = arctan(oo)$

Let $x = arctan(oo)$ .

$=> tanx=oo$ .

When we say that something is equal to infinity, in this context, it means that it approaches infinity only.

$sinx/cosx -> oo$

Now, we know that a quotient approaches infinity either if the numerator also approaches infinity or if the denominator approaches zero.

But the maximum value of $sinx$ is $1$ , thus $cosx$ must approach $0$ .

$cosx->0 => x->pi/2$

Usually, we would add a $+2npi$ at the end, but the co-domain of the inverse function is defined as $[0,2pi]$ , so we need not do it.

$x=pi/2 => arctan(oo) = pi/2 => color(red)(lim_(n->oo) arctan(n!)=pi/2$