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

Question

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

1859 views around the world

You can reuse this answer
Creative Commons License

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#