Limit exists proof

[william_33]

Show that if \(\displaystyle f_n\) js defined on \(\displaystyle R\) by \(\displaystyle f_n(x)=\frac{2}{\pi}\text{Arctan}(nx),\)

then \(\displaystyle f=lim(f_n)\) exists on \(\displaystyle R\).

How can I prove this problem?

 

[daon2]

Its a bounded monotone sequence...

 

[HallsofIvy]

Its a bounded monotone sequence...

Since this is a sequence of functions, what metric and ordering are you using?

 

[daon2]

Since this is a sequence of functions, what metric and ordering are you using?

The real metric. The function f depends only on the sign of x.

 

[william_33]

But I am confused on how to prove this? I know that it is monotoned and bounded.

 

[daon2]

But I am confused on how to prove this? I know that it is monotoned and bounded.

Since for a fixed \(\displaystyle x\), \(\displaystyle f_n(x)\) is a bounded an monotone sequence (the specific behavior will depend on the sign of x), the monotone convergence theorem tells you that it converges. So the sequence converges pointwise to some function. It is also easy to determine what it converges to with a little thought.