Direct substitution gives #0/0# so we can use L'Hopital's rule:
#lim_(x to 3)(3^x-27+ln(x-2))/(x-3) =#
#lim_(x to 3)(d/dx(3^x-27+ln(x-2)))/(d/dx(x-3)) =#
#lim_(x to 3)(3^x*ln(3)+(1/(x-2)))/(1)#
now we can substitute:
#lim_(x to 3)(3^x*ln(3)+(1/(x-2)))/(1) = 27ln(3)+1#
A second approach using just derivatives:
If you take the original limit and break it into two limits you have:
#lim_(x to 3)(3^x-27+ln(x-2))/(x-3) = #
#lim_(x to 3)(3^x-27)/(x-3)+lim_(x to 3)(ln(x-2))/(x-3)#
notice that #ln(3-2)=ln(1) = 0# so we can add one more detail to our new limits to get:
#lim_(x to 3)(3^x-27)/(x-3)+lim_(x to 3)(ln(x-2) - ln(3-2))/(x-3)#
Let's look at the left limit:
#lim_(x to 3)(3^x-27)/(x-3)#
This is the definition of the derivative of the function #f(x)=3^x# at #x=3#.
For #f(x)=3^x# we know #f'(x)=3^x*ln(3)#, so:
#lim_(x to 3)(3^x-27)/(x-3)=f'(3)=3^3ln(3)=27ln(3)#
For the right limit, it's the definition of the derivative of #f(x)=ln(x-2)# at #x=3#.
For #f(x)=ln(x-2)# we know that #f'(x)=1/(x-2)#, so:
#lim_(x to 3)(ln(x-2) - ln(3-2))/(x-3)=f'(3)=1/(3-2)=1#
So the overall limit will be the sum of the values of these two derivatives:
#lim_(x to 3)(3^x-27)/(x-3)+lim_(x to 3)(ln(x-2) - ln(3-2))/(x-3)=27ln(3)+1#
