# Thread:

1. ## limit problem

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 = 8.

is this the correct method of doing the problem?

in case it looked confusing up there, it should read: "the limit as x approaches 1 of (f(x) - 8) / (x - 1) equals 10."

2. Originally Posted by oblixps
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 = 8.

is this the correct method of doing the problem?

in case it looked confusing up there, it should read: "the limit as x approaches 1 of (f(x) - 8) / (x - 1) equals 10."
If you let $g(x)=\frac{f(x)-8}{x-1}$, we have $f(x)=g(x)(x-1)+8$.

Therefore, $\lim_{x\to1}f(x)=\lim_{x\to1}g(x)(x-1)+8=0+8=8$.

Does this make sense?
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