University Calculus problem help!!!?

Question

I have a University Calculus textbook and I have a question on one of the promblems of chapter 2 section 2. If you have the same textbook or are just smarter than me and know how to do this question help, please. It is number 85a. Here it is:

If limit as x goes to 2 is (f(x)-5)/(x-2)=3, find limit as f(x) goes to 2.

My question is how do you find the limit of a function with a function, f(x), in it? The book answer says 5, but I'm not sure how they got it. Thanks to anyone who answers in advance.

I think I just need to make the numuerator equal zero, so that with the limit of 2 I get 0/0.

Answer 1

as x goes to 2, x-2 goes to 0. If f(x)-5 doesnt go to 0 then the fractal (f(x)-5)/(x-2) goes to infinite. but this is not true because the fractal (f(x)-5)/(x-2) goes to 3. then it must f(x)-5 goes to 0 as x goes to 2, i.e f(x) goes to 5 as x goes to 2.

Answer 2

Notice that as x -> 2, the denominator -> 0. If f(x) - 5 had a nonzero limit, then (f(x) - 5)/(x - 2) would have limit +/- infinity. But since the quotient has a finite limit, the numerator must go to 0 as x -> 2. Therefore, lim(f(x)) = 5.

Answer 3

if the decrease of the sequence isn't 0, then the sequence diverges. decrease as n--> infinity. use the spinoff rule, L hosital's or something. spinoff of n = a million spinoff of ln(n) = a million/n a million / (a million/n) = n take decrease as n is going to infinity of n, and you get infinity. The decrease isn't 0, so it diverges.