Finding f(x) Using Limit and Algebraic Manipulation

[tg43fly]

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

 

[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 don't 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 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.

 

[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.

 

[tg43fly]

would lim x->1 g(x) = 8?

 

[johnqwertyful]

would lim x->1 g(x) = 8?

That depends, what's g(x)?

 

[tg43fly]

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?

 

[johnqwertyful]

Please stick to one problem. It will get confusing otherwise.

 

[tg43fly]

im too confused as with limit laws in general, ill find some more basic examples to start with. ty for helping.

 

[lendav_rott]

Whoah, I'm interested in this one too now.

If lim x->1 g(x) = c what do you mean by "what g(x) would simplify things" in the product of the 2 limits?

 

[johnqwertyful]

Whoah, I'm interested in this one too now.

If lim x->1 g(x) = c what do you mean by "what g(x) would simplify things" in the product of the 2 limits?

You CREATE a new limit to multiply with the limit you're given. What limit would simplify things?

 

[johnqwertyful]

Just create any g(x), find the limit, see if it simplifies things.

Try g(x)=cos(x). Try g(x)=x^2. Try g(x)=sqrt(x).
See what happens. Does it simplify things?

 

[Curious3141]

Why make the problem so complicated?

The ONLY way for the limit to exist is for the ratio to tend to the indeterminate form $\frac{0}{0}$ as $x \to 1$.

That happens when $\lim_{x \to 1} (f(x) - 8) = 0$, or simply $\lim_{x \to 1} f(x) = 8$.

That's all that's needed.

The RHS doesn't influence that answer at all. What it does is influence the limit of $f'(x)$ (the first derivative) as $x \to 1$. By L' Hopital's Rule, differentiating the numerator and denominator separately gives the result $\lim_{x \to 1} f'(x) = 10$. But this part is just for interest, and not needed for the original answer.

 

[johnqwertyful]

That's another way to do it, although I'm not sure if he cover l'hopitals.I would comment that you don't know f is differentiable; but if it is differentiable, your last paragraph is cool.

 

[Curious3141]

That's another way to do it, although I'm not sure if he cover l'hopitals.


I would comment that you don't know f is differentiable; but if it is differentiable, your last paragraph is cool.

There is no need to invoke L' Hopital's for the answer - that's just algebra.

I only used L' Hopital's Rule for the latter part ("for interest"), which is where the RHS would come into play. I should've stated that f(x) has to be differentiable around the neighbourhood of x = 1 for this to be valid.