lim x-> infinite x*[lnx - ln(x-3)]

Question

Working on my calc bc homework and mildly stumped by this problem. I know that when graphed the answer is 3, but I need to get it into a form where I can use L'Hopital's rule and work it out. 

Answer 1

x[lnx - ln(x-3)] = ln[x/(x-3]/ (1/x)

 

                     = ln[1/(1-3/x)]/(1/x)

 

                     = [ln1 - ln(1-3/x)]/(1/x)

 

                     = -ln(1-3/x)/(1/x)

 

As x → ∞, this expression → 0/0, so we can use L'Hopital's Rule.

 

Applying L'Hopital's Rule, lim [-ln(1-3/x)/(1/x)]  (as x → ∞)

 

                                  = lim [( (-3/x²)/(1-3/x) )/(-1/x²)] (as x → ∞)

 

                                  = lim[3/(1-3/x)] (as x→∞) = 3/(1-0) = 3