Finding the Limit of f(x) as x Approaches 1

[demonelite123]

lim [f(x) - 8] / [x - 1] = 10
x ~> 1

What is the limit as x approaches 1 of f(x)?
(Hint: let g(x) = [f(x) - 8] / [x - 1])

i'm not sure how the hint helps me in solving this problem. all i did was multiply both sides of the equation [f(x) - 8] / [x - 1] = 10 by (x - 1) and then added 8 to solve for f(x). then i got f(x) = 10x - 2 and i found that the limit of f(x) as x approaches 1 is 8. is that right? i just kinda guessed on how to do this problem. i don't really understand it.

 

[Masna]

I think you may have posted the limit incorrectly. The limit of all of that is equal to 10, as is, and must evaluated thusly. Is the limit simply the limit of f(x) as x goes to 1? Then, maybe, the problem states that that limit minus 8 divided by (x - 1) is equal to 10?

 

[jbunniii]

lim [f(x) - 8] / [x - 1] = 10
x ~> 1

What is the limit as x approaches 1 of f(x)?
(Hint: let g(x) = [f(x) - 8] / [x - 1])

i'm not sure how the hint helps me in solving this problem. all i did was multiply both sides of the equation [f(x) - 8] / [x - 1] = 10 by (x - 1) and then added 8 to solve for f(x). then i got f(x) = 10x - 2 and i found that the limit of f(x) as x approaches 1 is 8. is that right? i just kinda guessed on how to do this problem. i don't really understand it.

Your final answer is right, but your method is wrong. The problem is that your method implicitly assumed that [f(x) - 8] / [x - 1] = 10 is true for all x close to 1 but not equal to 1 (thus allowing you to evaluate the limit). But it might not be true for any x at all!

Think about it this way: as x -> 1, the denominator goes to zero. Yet the quotient approaches a finite number. This can happen ONLY if the numerator also goes to zero. Therefore you must have f(x) -> 8 as x -> 1. You can argue this more formally with epsilons and deltas, but that's the basic idea.

Note that you would get the same answer if "10" was replaced by any other finite number. All you need is the finiteness of the right hand side. If you know L'Hopital's rule, you can see that the right hand side is controlling f'(1), the derivative of f at x=1, assuming it exists.