#lim_(n->oo)(1+xe^(nx))/(1+e^(nx));x inRR#?

2 Answers
Cesareo R.
Jul 11, 2017

See below.

Explanation:

#lim_(n-> oo)(1+xe^(nx))/(1+e^(nx)) = {(x < 0-> 1),(x=0->1/2),(x > 0->x):}#

Truong-Son N.
Jul 11, 2017

Well, we could split this into three parts:

  • #x > 0#
  • #x = 0#
  • #x < 0#

If #x > 0#, then

#L = lim_(n->oo) (1 + xe^(nx))/(1 + e^(nx))#

At large #n#, #1# is small compared to #e^(nx)#. So:

#=> color(blue)(L) = lim_(n->oo) (xcancel(e^(nx)))/(cancel(e^(nx))) = color(blue)(x)#

If #x < 0#, then at large #n#, if we denote "#x#" as positive, then #e^(-nx) -> 0#, so:

#=> color(blue)(L) = lim_(n->oo) (1 + xe^(-nx))/(1 + e^(-nx))#

#= lim_(n->oo) (1 + cancel(x/(e^(nx))))/(1 + cancel(1/(e^(nx)))) = color(blue)(1)#

If #x = 0#, then

#L = lim_(n->oo) (1 + xe^(nx))/(1 + e^(nx))# is of the form #(0cdote^(oocdot0))/(e^(oocdot0))#.

It seems odd at first, since #e^(oocdot0)# is undefined, but #e^(nx) -> 1# at #x = 0#, even when #n -> oo#, by looking at the graph of #y = e^(nx)# for large #n#.

So,

#=> color(blue)(L) = (1 + 0cdot1)/(1 + 1) = color(blue)(1/2)#

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