#lim_(x->0)((3(e^x-1))-3x)/(x(e^x-1))#
If we plug #0# directly into our equation, we get the outcome:
#lim_(x->0)((3(e^x-1))-3x)/(x(e^x-1))# "=" #0/0#
(Quotations to ensure I'm not stating that the two are equal and is actually undefined)
However, this allows us to use L'Hospital's Rule.
We need to have #0/0# or #oo/oo# for L'Hospital's Rule to be applicable.
Thus, we can move forward with that rule, but lets first distribute to help up with taking derivatives:
#lim_(x->0)((3(e^x-1))-3x)/(x(e^x-1)) = lim_(x->0)(3e^x-3x-3)/(xe^x-x)#
#=>lim_(x->0)(d/dx[3e^x-3x-3])/(d/dx[xe^x-x]) = lim_(x->0)(3e^x-3)/(e^x+xe^x-1#
Now, let's try plugging in #0# for #x#:
#lim_(x->0)(3e^x-3)/(e^x+xe^x-1) = (3e^0-3)/(e^0-0e^0-1)#"="#0/0#
The answer is still undefined, but we can apply L'Hospital's Rule once again since we got an output of #0/0#.
#lim_(x->0)(3e^x-3)/(e^x+xe^x-1) = lim_(x->0)[d/dx(3e^x-3)]/[d/dx(e^x+xe^x-1)]#
#=lim_(x->0)(3e^x)/(2e^x+xe^x) = (3e^0)/(e^0+0e^0-1) = 3/2#
Thus,
#lim_(x->0)((3(e^x-1))-3x)/(x(e^x-1))=3/2#
