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

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### 1 Answer

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**