# lim(x->0)((1-e^2x)/(1-e^x))

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`lim_(x->0)[1-e^(2x)]/[1-e^x]=`

`lim_(x->0)[(1-e^x)(1+e^x)]/(1-e^x)=`

`lim_(x->0)1+e^x=`

`1+1=2`

otner solution:

By L'Hôpital's rule we get:

lim as x goes to 0 ( -2e^2x / - e^x ) =

lim as x goes to 0 ( 2e^x ) = 2