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
