A wierd limit question (1 Viewer)

[sasquatch]

This question as i know is very simple but theres something i dont get..

y = (x² - x + 1) / (x-1)

Finding the horizontal asymptote:

y = lim x-> inf (x²/x - x/x + 1/x) / (x/x-1/x)​

= lim x-> inf (x - 1 + 1/x) / (1-1/x)​

= (x - 1 + 0) / (1-0)​

= x -1​

But if you were the change the original equation to:

y = (x² - x) / (x-1) + 1 / (x-1)​

= x(x -1) / (x-1) + 1/(x-1)​

= x + 1/(x-1)​

and then performing the limit..

y = lim x-> inf x + (1/x) / (x/x -1/x)​

= lim x-> inf x + (/1x) / (1- 1/x)​

= x + 0/1​

= x​

How does this work out?? One says x - 1 and the other x, when it is the same equation??

 

[jake2.0]

y=x is the oblique asymptote and x=1 is the vertical asymptote

 

[who_loves_maths]

Hi sasquatch,

to ensure you obtain the correct answer in these limit questions, the correct procedure is to factorise and cancel first before you substitute the limits in. this is because of additional factors which can cause confusion.
your second method in which you obtained y = x as the asymptote is the correct method.

the issue you raise is actually very complex and goes to the heart of the definition of infinity and orders of infinity.

but to simplify matters here, look at it this way:
in your first method, you divided by 'x' top and bottom. but let me ask you this... what stopped you from dividing by x^2 ? why did you choose 'x' preferentially over 'x^2'... there should be no reason.
if we instead divided by x^2, you get:
lim (1 - 1/x + 1/x^2)/(1/x - 1/x^2) ---> 1/0 ---> infinity for x -> infinity.
here, you do not get any 'x's, and only end up with the value of "infinity".

but, if you think about it, your method of dividing by 'x' is similar:
you got y = x - 1, but if x -> infinity, then y -> infinity -1 -> infinity
which is the same as the method of dividing by 'x^2' even though you initially get different answers.

this is to illustrate to you that the definition of infinity is the problem here.
that is: infinity -1 = infinity

just as you say y = x - 1 in your method, i can say y -> infinity as x -> infinity, so if "infinity" is one value denoted by "a", then as y -> a, x -> a also.
hence, the asymptote is y = x for large values of 'x'.

the thing is, when dealing with large values of 'x', (x - 1) and (x) differ by an infinitesimal amount! that's why you end up with both answer when substituting the limit x -> infinity in both your methods.
but this is inconsistent, because in your first method, you left the 'x' alone and got rid of the 1/x and 1/x^2. yet had you been consistent with your substitution and put in x = infinity, you would have obtained y = infinity -1 = infinity ---> ie. y = x is asymptote.

... i know i'm not very clear in my explanation, but it's late, and like i said, the issue over the definition of "infinity" is a contentious issue.

just remember that the optimum method to use for these questions is to factorise fully first until the denominator is of a higher order polynomial than the numerator before substituting in limits.


eg. to illustrate my point, here's an alternative "trick" you can use on your second method:

y = (x^2 - x) / (x-1) + 1 / (x-1)
= x(x -1) / (x-1) + 1/(x-1)

...continued from here... factorise:
y = (x-1)[x/(x-1) + 1/(x-1)^2]

for x -> infinity, 1/(x-1)^2 -> 0 ; and x/x-1 -> 1

ie. y -> (x-1)[1 + 0] -> (x-1) is asymptote.

so from your second method, it seems we can obtain the asymptote y = x -1 as well?!

...

 

[who_loves_maths]

oh and i just realised - one more thing wrong with your method that adds to your confusion:

when seeking asymptotes, we do not use limits explicitly.

we use limits only for, say, horizontal asymptote because it gives up a definite value for 'y'. but in your case, the asymptote is oblique - and you don't use the limit notation/technique to look for oblique asymptotes lol!
this is exactly why it didn't work.
because as a limit, you should have gotten a value - independent of 'x's.
ie. y -> (infinity -1) -> (infinity) for your first method.
and y -> (infinity) for your second method.

ie. when finding limits, both your methods are valid.

but, when finding equations of oblique asymptotes, then you must factorise the equation of the asymptote out first!
why? because IF they exist, then their equations can always be factored out!

this is why your second method is the correct one for finding asymptotes (and not limiting values).

 

[NT-social]

your mistake is that you divided by x but ur meant to divide it by the one with highest degree...so ur meant to divide by x squared

 

[Riviet]

Yeah, the first thing you should do is divide by the highest power of x, this should only be used for finding horizontal asymptotes.

 

[sasquatch]

No i get slightly what you were saying who_loves_maths but the rest of you saying dividing by the highest power of x is wrong. Your suppoesd to divide by the highest power of x on the denominator!

Looky here.

y = (x² - x + 1) / (x-1)

y = lim x-> inf (x²/x² - x/x² + 1/x²) / (x/x² - 1/x²)

= lim x-> inf (1 - 1/x + 1/x²) / (1/x - 1/x²)
= (1 - 0 + 0) / (0 - 0)
= 1 / 0

which is undefined...so yeah...

 

[icycloud]

Because the numerator is of a higher degree than the denominator, it is a GIVEN that:

lim (x-->infinity) f(x) = infinity

Thus, there are no horizontal asymptotes.

 

[Trebla]

This question as i know is very simple but theres something i dont get..

y = (x² - x + 1) / (x-1)

Finding the horizontal asymptote:

y = lim x-> inf (x²/x - x/x + 1/x) / (x/x-1/x)​

= lim x-> inf (x - 1 + 1/x) / (1-1/x)​

= (x - 1 + 0) / (1-0)​

= x -1​

But if you were the change the original equation to:

y = (x² - x) / (x-1) + 1 / (x-1)​

= x(x -1) / (x-1) + 1/(x-1)​

= x + 1/(x-1)​

and then performing the limit..

y = lim x-> inf x + (1/x) / (x/x -1/x)​

= lim x-> inf x + (/1x) / (1- 1/x)​

= x + 0/1​

= x​

How does this work out?? One says x - 1 and the other x, when it is the same equation??

I think you did something wrong.
When you sub in the value x = infinity don't you have to sub it in for ALL x? What you have done is just sub in x = infinity into the denominators and left the x alone. With limits I don't think you can do that.
For example when you ended up with:
lim (x --> ∞) (x - 1 + 1/x) / (1-1/x)
When you take the limit you should end up with:
(∞ - 1 + 1/∞) / (1 - 1/∞) = (∞ - 1 + 0) / (1 - 0)
= ∞ - 1
I think that's where your problem is.
Same goes for this:
lim (x --> ∞) x + (1/x) / (x/x -1/x)
When you take the limit you should end up with:
∞ + (1/∞) / (1 - 1/∞) = ∞

If this is a limit question there should be no definite limit because it would be something like x = ∞
As for asymptotes, as far as I can tell there aren't any horizontal asymptotes other than something like x = ∞

 

[sasquatch]

Um i dont know what your on about.

The answer actually for the asymptote is y = x, as i made some graphing program and yeah its pretty obvious to see.

Also there is a maximum turning point i think at (0,-1) and if y = x - 1 for the asymptote then

-1 = 0 -1
-1 = -1

the point (0,-1) lies on the asymptote y = x - 1 which is impossible as it would not be an asymptote... well i think thats the case for oblique asymptotes. I know that points can lie on horizontal asymptotes but........ yeah i dunno...

 

[Trebla]

Here's what I mean:
Simply take the following example:
lim (x --> 2) x(x² - 4)/(x - 2)
Simplify:
lim (x --> 2) x(x - 2)(x + 2) / (x - 2)
= lim (x --> 2) x(x + 2)
Take the limit:
= 2(2 + 2) = 8
What YOU did was something like this:
= lim (x --> 2) x(x + 2)
= x(2 + 2)
= 4x
You forgot the extra x. There shouldn't be any x values when you take the limit.
You forgot to substitute for ALL x in your question. Isn't that what you do in limits?
Either way, even if you did find the asymptotes by inspection you have to show how you worked it out. In a scenario like this it would often by long division.

 

[acmilan]

Here's what I mean:
Simply take the following example:
lim (x --> 2) x(x² - 4)/(x - 2)
Simplify:
lim (x --> 2) x(x - 2)(x + 2) / (x - 2)
= lim (x --> 2) x(x + 2)
Take the limit:
= 2(2 + 2) = 8
What YOU did was something like this:
= lim (x --> 2) x(x + 2)
= x(2 + 2)
= 4x
You forgot the extra x. There shouldn't be any x values when you take the limit.
You forgot to substitute for ALL x in your question. Isn't that what you do in limits?
Either way, even if you did find the asymptotes by inspection you have to show how you worked it out. In a scenario like this it would often by long division.

When you use the long division method you're effectively using the same method above which you said was wrong.

For example, say you have a function f(x), which, after long division, becomes f(x) = x + 1/x

Now you consider the value as x goes to infinity. In this case, the 1/x becomes insignificant, 'sub' in infinity and it goes to 0, and you're left with mainly x, the limiting value of the curve tends to the limiting value of y = x. Yes, if you take the limiting value you aren't supposed to have any x values, but you arent concerned with what the limiting value is, your aim is to remove the values of x which become insignificant as x becomes large.

A good way I think you can do it (using my example) is like this:

lim (x->∞) f(x)
= lim (x->∞) x + 1/x
= lim (x->∞) x + lim (x->∞) 1/x
= lim (x->∞) x + 0

.: lim (x->∞) f(x) = lim (x->∞) x

Which shows as x becomes large, f(x) tends to the line y = x

 

[sasquatch]

Yeah i dont know what the hell you were on about Trebla and if you saw i did show that the assymptote was y = x... just that using the original equation i got y = x - 1.. that was the whole problem....

Oh and also check my working i did not forget to subsitute for all values of x.

 

[who_loves_maths]

Originally Posted by sasquatch
No i get slightly what you were saying who_loves_maths but ...

hi again sasquatch,

okay, to put it as simple as possible, the gist of what i said is that you get different answers with different methods because you are inconsistent in your application of the limit to find asymptotes:

1) there is a difference between finding a limiting value (limit) and finding an asymptote - they are not the same thing, and the procedure for finding them are subtly different.

2) when finding a limiting value, use the substitution to the fullest extent - do not substitute x = infinity for certain values but not others. this "preferential discrimination" between the 'x's in the equation is an inconsistency that gives rise to your dilemma between two different methods.

ie. when finding the limiting value, both your answers yield the same consistent value (as they should). since (infinity - 1) = (infinity)

3) having dealt with finding limiting values, let's talk about finding asymptotes.
an asymptote exists where as x -> infinity, y -> the asymptote (equation).
hence, an asymptote can be defined as:
y = f(x) + F(infinity) = f(x) + 0
where f(x) is the asymptote function, and F(x) is the remnant function such that F(infinity) = 0
this allows us to arrive at the asymptote function: y = f(x) .

so now, having defined that clearly, we can look at the implications of it. from the definition, it's quite obvious now that in order to fix the asymptote function, you need first to look for the remnant function F(x) such that F(infinity) = 0.
so let's apply that to your rational function:

y = g(x) = (x2 - x + 1) / (x-1) ; now since deg(numerator) > deg(denominator), then g(x) itself is not the remnant function since g(infinity) would not tend to 0.
so the next question is: how do we find the remnant function?
easy. for all rational functions, the only method of separating two functions f(x) and F(x) is via partitioning of g(x).
ie. most usually via polynomial division.

this now explains why your second method is the valid one - since you applied polynomial division to simplify g(x) into:
g(x) = x + 1/(x-1)
here, F(x) = 1/(x-1) where F(infinity) = 0, since deg(numerator) < deg(denominator) for F(x).
and, f(x) = x

ie. y = x + 1/(x-1) = f(x) + F(x)
---> y = f(x) = x is thus the asymptote by our definition.


hope that clears some things up for you . type back if you don't see what i'm trying to say.