Finding f(x) Using Limit and Algebraic Manipulation

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Homework Help Overview

The discussion revolves around finding the function f(x) given the limit condition lim x->1 (f(x)-8)/(x-1) = 10. Participants are exploring the implications of this limit and the behavior of f(x) near x = 1.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the correct interpretation of the limit expression and question the assumptions about the function f(x). There are attempts to manipulate the limit and explore potential forms for f(x) and g(x) that could simplify the problem.

Discussion Status

The discussion is ongoing, with various interpretations and approaches being explored. Some participants suggest creating a new function g(x) to facilitate the limit calculation, while others express confusion about the relationship between f(x) and the limit condition. There is no explicit consensus on the next steps or the form of f(x).

Contextual Notes

Participants note the lack of additional information about f(x) and the constraints imposed by the limit condition. There is mention of the indeterminate form that arises as x approaches 1, which is central to the discussion.

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Homework Statement



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

Homework Equations





The Attempt at a Solution


i don't know where to start on finding f(x), i assume it includes (x-1) to eliminate the 0 denominator, i need some hints
 
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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.
 
yeah sorry, it is indeed the latter
lim x->1 (f(x)-8)/(x-1) = 10
 
Last edited:
But even then, you can only know what the limit is at one point. You know nothing of the function. Is there more information?
 
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 don't think that worked
 
Is the problem

find f(x)

or is the problem

find lim x-> 1 f(x)
 
its find lim x-> 1 f(x)
 
So you can multiply limits, right? So you can multiply the given limit by something. What something would make it easier?
 
i don't comprehend...
do you mean bringing (x-1) to above by inversing it? i tried that.
the whole f(x) within another function boggles me.
 
  • #10
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?
 
  • #11
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)?
 
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  • #12
"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?
 
  • #13
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?
 
  • #14
Huh? Would what address what? I think you might be overthinking this.
 
  • #15
tg43fly said:
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.
 
  • #16
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
 
  • #17
Isn't that the limit you're given?
 
  • #18
Also, you can't factor anything. x-1 is irreducible.
 
  • #19
i'm so lost.
does that mean f(x) can just be (x-1) to factor out the denominator?
 
  • #20
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.
 
  • #21
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.).
 
  • #22
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)?
 
  • #23
lim x->1 (f(x)-8)(g(x)) / (x-1) = 10*g(x)
so i have to find g(x) which is a constant?
 
  • #24
tg43fly said:
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)
 
  • #25
eq0005MP.gif


This is what I'm using.
 
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  • #26
would lim x->1 g(x) = 8?
 
  • #27
tg43fly said:
would lim x->1 g(x) = 8?

That depends, what's g(x)?
 
  • #28
i'm looking at an similar example lim x->5 f(x) / (g(x) + 4)
however they've given the values of lim x-> 5 f(x) = 2 and g(x) = -4 albeit this function does not limit
in my case how would i determine g(x)?

can i rewrite g(x) into a polynomial function that gives a constant c given lim x-> 1?
 
  • #29
Please stick to one problem. It will get confusing otherwise.
 
  • #30
im too confused as with limit laws in general, ill find some more basic examples to start with. ty for helping.
 

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