Question

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

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

Answer 1

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

Explanation:

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`