`lim_(x->oo) e^(-x) = 0`

So:

`lim_(x-> oo) (1+2e^(-x))/(1-e^(-x)) = (1+2*0)/(1-0) = 1`

`lim_(x->-oo) e^(-x) = lim_(x->oo)e^x = oo`

So: `lim_(x->-oo) (1+2e^(-x))/(1-e^(-x)) = (oo)/(oo)`

This is an indeterminate form, and so we need to use L'hopital's rule. We take the derivative of both the numerator and the denominator, and...

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`lim_(x->oo) e^(-x) = 0`

So:

`lim_(x-> oo) (1+2e^(-x))/(1-e^(-x)) = (1+2*0)/(1-0) = 1`

`lim_(x->-oo) e^(-x) = lim_(x->oo)e^x = oo`

So: `lim_(x->-oo) (1+2e^(-x))/(1-e^(-x)) = (oo)/(oo)`

This is an indeterminate form, and so we need to use L'hopital's rule. We take the derivative of both the numerator and the denominator, and then take the limit:

`lim_(x->-oo) (-2e^(-x))/(e^(-x)) = lim_(x->-oo) -2 = -2`