Evaluate the limit of f(x)=[1+1/(x^2-2x)]^2x

Since it is not specified if x tends to a certain value,
we'll evaluate the classical case when x tends to
infinite.

We notice that if x->infinite, we'll get
the case of indeterminacy 1^infinite.

We'll create the
elementary limit e.

For this reason, we'll multiply the
suprescript 2x by x^2-2x and we'll divide by the same amount
x^2-2x.

Lim f(x) = lim
{[1+1/(x^2-2x)]^x^2-2x}^2x/(x^2-2x)

We know that lim
[1+1/u(x)]^u(x) = e, if u(x)-> infinite

lim
{[1+1/(x^2-2x)]^x^2-2x}^2x/(x^2-2x) = e^lim
2x/(x^2-2x)

e^lim 2x/(x^2-2x) = e^lim 2x/x^2(1 - 2/x) = e^0
= 1

lim [1+1/(x^2-2x)]^2x = 1, if x->
infinite