Finding f(x) given limit

[tg43fly]

Homework Statement

lin x-> 1
f(x)-8 / x-1 = 10
find f(x)

Homework Equations

The Attempt at a Solution

i dont know where to start on finding f(x), i assume it includes (x-1) to eliminate the 0 denominator, i need some hints

 

[johnqwertyful]

Please include parenthesis. Is it f(x)-8/x-1 or is it (f(x)-8)/(x-1). I'm assuming the latter. Also what is the given limit?

Edit, I see the limit now. Your notation was confusing. To answer your question, you know you can multiply limits, right? So lim x-> 1 (f(x)-8)/(x-1)=10, then you can multiply by a limit that you KNOW exists.

 

[tg43fly]

yeah sorry, it is indeed the latter
lim x->1 (f(x)-8)/(x-1) = 10

 

[johnqwertyful]

But even then, you can only know what the limit is at one point. You know nothing of the function. Is there more information?

 

[tg43fly]

no, that was all that given; find f(x) at lim x-> 1
any idea where i should start, i did trial and error on lim x->1, f(x) = (x-1)^n, but i dont think that worked

 

[johnqwertyful]

Is the problem

find f(x)

or is the problem

find lim x-> 1 f(x)

 

[tg43fly]

its find lim x-> 1 f(x)

 

[johnqwertyful]

So you can multiply limits, right? So you can multiply the given limit by something. What something would make it easier?

 

[tg43fly]

i dont comprehend...
do you mean bringing (x-1) to above by inversing it? i tried that.
the whole f(x) within another function boggles me.

 

[johnqwertyful]

That's not what I meant at all. You can create an entirely NEW limit, then multiply it with your given limit, so long as the limit you create exists. What would simplify things?

 

[tg43fly]

so i.e.
lim x->1 -8/(x-1) * f(x) where i can make up a new limit for f(x) which will also need (x-1)? do i need to get rid of lim x->1 (x-1) on the denominator with the f(x)?

 

[johnqwertyful]

"Make up a new limit for f(x)". No, you cannot "make up" a limit for f(x). That limit just is. But you CAN create a new function, g(x), such that lim x-> 1 g(x)=c, some constant (you create g(x), and from this you know what the constant is)

So then lim x-> 1 (f(x)-8)*g(x)/(x-1)=10*c

What g(x) would simplify this?

 

[tg43fly]

would this also address the 0 denominator given lim x->1 for (x-1)?
f(x) * g(x) = 10x-2
would this be on the right path?

 

[johnqwertyful]

Huh? Would what address what? I think you might be overthinking this.

 

[johnqwertyful]

would this also address the 0 denominator given lim x->1 for (x-1)?
f(x) * g(x) = 10x-2
would this be on the right path?

I have no idea where that equation came from.

 

[tg43fly]

does (f(x) - 8) / (x - 1) = 10 need to hold true for lim x->1? i thought id have to factor out the x-1 since it'd give a 0 denominator

 

[johnqwertyful]

Isn't that the limit you're given?

 

[johnqwertyful]

Also, you can't factor anything. x-1 is irreducible.

 

[tg43fly]

i'm so lost.
does that mean f(x) can just be (x-1) to factor out the denominator?

 

[johnqwertyful]

There is no point in this problem where you will factor anything. Also, there is no point in the problem where you will ever know what f(x) is.

Like I said, you need to multiply the limit you are given by another limit that you create.

 

[johnqwertyful]

I'll go one step further. FIRST you must multiply the limit you are given by another limit you create. Then you must add another limit to it. THEN you will be done. The tricky part is coming up with those two limits (but it'll click and make sense after the first one.).

 

[johnqwertyful]

So then lim x-> 1 (f(x)-8)*g(x)/(x-1)=10*c

What g(x) would simplify this?

This is the only thing you should be thinking about right now. What g(x) will make this easier? It won't solve it, there is another step coming, but what will simplify it? Also, what is the associated "c", the lim x->1 g(x)?

 

[tg43fly]

lim x->1 (f(x)-8)(g(x)) / (x-1) = 10*g(x)
so i have to find g(x) which is a constant?

 

[johnqwertyful]

lim x->1 (f(x)-8)(g(x)) / (x-1) = 10*g(x)
so i have to find g(x) which is a constant?

g(x) is a function. c is the limit as x->1 of g(x)

 

[johnqwertyful]

This is what I'm using.