1. ## LIMITS again

Hello

Heres the problem:

It is know that f and g are defined for every x.
It is known that

I am asked to prove or disprove the following statement:

Now i think you can just disprove it when both f(x) and g(x) are approaching 0 when x -> 1 then you get 0/0 which isnt = 1.
(Just make up some random functions that would do that like f(x) = x-1, and g(x) = 2x-2)

you'd get x-1/x-1.

The teacher says otherwise but he didn't explained himself too well, something about "it is possible to algebraically manipulate the function to make the 0 disappear from the denominator."
It is possible in some functions alright but the functions aren't given here and we just need to disprove the statement so would that "disprovement" be correct?

Thanks a lot!

2. ## Re: LIMITS again

Originally Posted by ryu1
Heres the problem:
It is know that f and g are defined for every x.
It is known that
I am asked to prove or disprove the following statement:

Now i think you can just disprove it when both f(x) and g(x) are approaching 0 when x -> 1 then you get 0/0 which isnt = 1.
(Just make up some random functions that would do that like f(x) = x-1, and g(x) = 2x-2
you'd get x-1/x-1.

The teacher says otherwise but he didn't explained himself too well,
In this case he is wrong.

$\displaystyle {\lim _{x \to 1}}({x} - 1) = 0\;\& \;{\lim _{x \to 1}}(2x - 2) = 0$ but $\displaystyle {\lim _{x \to 1}}\frac{{{x} - 1}}{{2x - 2}} = \frac{1}{2}$

3. ## Re: LIMITS again

Originally Posted by ryu1
Hello

Heres the problem:

It is know that f and g are defined for every x.
It is known that

I am asked to prove or disprove the following statement:

Now i think you can just disprove it when both f(x) and g(x) are approaching 0 when x -> 1 then you get 0/0 which isnt = 1.
(Just make up some random functions that would do that like f(x) = x-1, and g(x) = 2x-2)

you'd get x-1/x-1.

The teacher says otherwise but he didn't explained himself too well, something about "it is possible to algebraically manipulate the function to make the 0 disappear from the denominator."
It is possible in some functions alright but the functions aren't given here and we just need to disprove the statement so would that "disprovement" be correct?

Thanks a lot!
Hi ryu1!

You gave a very good example with f(x) = x-1, and g(x) = 2x-2.

$\displaystyle \lim_{x \to 1} \frac {f(x)}{g(x)} = \lim_{x \to 1} \frac {x-1}{2x-2} = \lim_{x \to 1} \frac {(x-1)}{2(x-1)}$

Now your teacher was also right.
You can algebraically manipulate this to make the 0 disappear from the denominator.
Just divide both numerator and denominator by (x-1).
This is allowed for any $\displaystyle x \ne 1$ and this is what a limit is for: to predict where the value is going to from its context, even if it is not defined at 1.

The result is:

$\displaystyle \lim_{x \to 1} \frac {f(x)}{g(x)} = \lim_{x \to 1} \frac {(x-1)}{2(x-1)}= \lim_{x \to 1} \frac {1}{2} = \frac 1 2$

In particular this disproves the statement $\displaystyle \lim_{x \to 1} \frac {f(x)}{g(x)}=1$.

4. ## Re: LIMITS again

Originally Posted by Plato
In this case he is wrong.

$\displaystyle {\lim _{x \to 1}}({x} - 1) = 0\;\& \;{\lim _{x \to 1}}(2x - 2) = 0$ but $\displaystyle {\lim _{x \to 1}}\frac{{{x} - 1}}{{2x - 2}} = \frac{1}{2}$
Why it's not 0/0 from your example? you "could" first calculate the limit of the numerator and get 0 and then calculate the limit of the denominator and get 0.
how is it possible to first place the functions themselves and then calculate the limit of that whole expression at once? (which seems right as well)

isn't there a law in limits that allows to "split" the limit operation to the numerator and the denominator (calculating them first and then doing the division)?

Thanks a lot!

5. ## Re: LIMITS again

Originally Posted by ILikeSerena
Hi ryu1!

You gave a very good example with f(x) = x-1, and g(x) = 2x-2.

$\displaystyle \lim_{x \to 1} \frac {f(x)}{g(x)} = \lim_{x \to 1} \frac {x-1}{2x-2} = \lim_{x \to 1} \frac {(x-1)}{2(x-1)}$

Now your teacher was also right.
You can algebraically manipulate this to make the 0 disappear from the denominator.
Just divide both numerator and denominator by (x-1).
This is allowed for any $\displaystyle x \ne 1$ and this is what a limit is for: to predict where the value is going to from its context, even if it is not defined at 1.

The result is:

$\displaystyle \lim_{x \to 1} \frac {f(x)}{g(x)} = \lim_{x \to 1} \frac {(x-1)}{2(x-1)}= \lim_{x \to 1} \frac {1}{2} = \frac 1 2$

In particular this disproves the statement $\displaystyle \lim_{x \to 1} \frac {f(x)}{g(x)}=1$.
Oh yes now I see, that's what Plato did as well, but doesn't this assumes that x is not = 1?
From the instructions it is given the functions are defined for all x's.
does this condition changes anything in this disprovment(disproval ?)?

6. ## Re: LIMITS again

Originally Posted by ryu1
Oh yes now I see, that's what Plato did as well, but doesn't this assumes that x is not = 1?
From the instructions it is given the functions are defined for all x's.
does this condition changes anything in this disprovment(disproval ?)?
Yes, it does assume that $\displaystyle x \ne 1$.
It assumes x gets as close to 1 as possible, without ever really reaching it.
As a result the limit-expression gets as close to its final value as possible, again without ever really reaching it.

Since you can find an example where the statement is not true, the statement is disproved.
This is a proof by counter example or contradiction.

7. ## Re: LIMITS again

Originally Posted by ILikeSerena
Yes, it does assume that $\displaystyle x \ne 1$.
It assumes x gets as close to 1 as possible, without ever really reaching it.
As a result the limit-expression gets as close to its final value as possible, again without ever really reaching it.

Since you can find an example where the statement is not true, the statement is disproved.
This is a proof by counter example or contradiction.
OK I can accept that the functions themselves are still defined at x=1 nonetheless(?)
Another question, why can you get 1/2 OR 0/0?
If you'd first split the limit to find the limit of the numerator and the limit of the denominator and then find their quotient you'd get 0/0.
You can do that with limits? so you can get 2 different answers to the same problem?

Thanks again.

8. ## Re: LIMITS again

Originally Posted by ryu1
OK I can accept that the functions themselves are still defined at x=1 nonetheless(?)
Another question, why can you get 1/2 OR 0/0?
If you'd first split the limit to find the limit of the numerator and the limit of the denominator and then find their quotient you'd get 0/0.
You can do that with limits? so you can get 2 different answers to the same problem?

Thanks again.
No, you can't get 2 answers to a limit.
You can only split the limit in numerator and denominator if the result is defined.
The value 0/0 is not defined - it can be anything.
That's why we use a limit to find out what it really is, for specific functions f(x) and g(x).

9. ## Re: LIMITS again

I suggest that you explore this question with the added condition
$\displaystyle {\lim _{x \to 1}}f(x) = {\lim _{x \to 1}}g(x) \ne 0$.

You should find the conclusion valid then.

10. ## Re: LIMITS again

I still struggle with a basic order of things with limits that relates to this questions.

What is the order of operations?
when you have do you first take the limit of f(x) and g(x) separately and THEN divide them(the limits)?
OR do you first evaluate f(x)/g(x) and take the limit of the result?

Thanks again.

11. ## Re: LIMITS again

Originally Posted by ryu1
I still struggle with a basic order of things with limits that relates to this questions.
What is the order of operations?
when you have do you first take the limit of f(x) and g(x) separately and THEN divide them(the limits)? OR do you first evaluate f(x)/g(x) and take the limit of the result?
Here is the theorem: If each of $\displaystyle {\lim _{x \to a}}f(x)\;\& \;{\lim _{x \to a}}g(x)$ exists and $\displaystyle {\lim _{x \to a}}g(x)\ne0$ then $\displaystyle \displaystyle{\lim _{x \to a}}\frac{{f(x)}}{{g(x)}} = \frac{{{{\displaystyle\lim }_{x \to a}}f(x)}}{{{{\displaystyle\lim }_{x \to a}}g(x)}}$

But it may be that we get $\displaystyle \frac{0}{0}$ in which case other algebraic operations are tried.

12. ## Re: LIMITS again

1. But which do you do first? the limit operation or division operation?

2. Can you say you will always be able to change the 0 in the denominator in such a case? (using algebra)
thanks.

13. ## Re: LIMITS again

Originally Posted by ryu1
1. But which do you do first? the limit operation or division operation?
2. Can you say you will always be able to change the 0 in the denominator in such a case? (using algebra)
thanks.
There is no way to answer #1 other than the theorem I posted . Use it!

For #2, here are two examples.

$\displaystyle {\lim _{x \to 1}}\frac{{{x^2} + 3x - 4}}{{{x^2} - 1}} = \frac{5}{2}$ and $\displaystyle {\lim _{x \to 1}}\frac{{\sqrt {x + 3} - 2}}{{x - 1}} = \frac{1}{4}$.