L'hopital's rule

Question

Hi, please help me find the solution to this question

lim. ( f(x) - 6) / (x-1)

x->1

Answer 1

Because you are using L'Hospitals rule, the assumption is that f(x) - 6 = 0; the reason being that L'Hospital's rule only works with the forms infinity over infinity or 0 over 0; because the bottom equals zero this means that the top must also be 0 for L'Hospital's rule to work. With all of that in mind f(1) must = 6 for L'Hospitals method to work. In order to use the rule you take the derivative of the top and the bottom. Start with the bottom because it is easier. The derivative of (x -1) is equal to 1; The derivative of f(x) - 6 is equal to f'(x); What we are left with is the Lim x-> 1 of (f'(x))/1; this equation no longer takes the form 0 over 0 of infinity over infinity so L'Hospital is no longer required. This means that given the information we have the best answer that can be deduced based the above assumptions is that Lim x->1 (f(x)-6)/(x-1) = f'(1)

Answer 2

We need to know what f(x) is in order to completely answer the question. (Note Eli's comment.) By L'Hopital's rule,

lim_x→1 {[f(x) - 6] / (x - 1)} = lim_x→1 f'(x)

But without knowing f(x), we cannot know f'(x). We need more information.