Prove or Disprove: Limit of Difference Equals Zero

[Torshi]

Homework Statement

Prove following statement is true or find a counterexample to demonstrate it is false?

Homework Equations

If lim x->3 f(x) / g(x) = 1 then lim x->3 (f(x)-g(x)) = 0

The Attempt at a Solution

well 3-3 = 0 and 3/3 = 1/1 = 1? But I think it's more than that

 

[jbunniii]

Can you find an example where $\lim_{x \rightarrow 3^+} \frac{f(x)}{g(x)} = 1$ but $\lim_{x \rightarrow 3^+} f(x)$ and $\lim_{x \rightarrow 3^+} g(x)$ do not exist?

 

[Torshi]

sorry - I would like to fix the limit! I meant lim x->3 without the "+"

 

[Dick]

sorry - I would like to fix the limit! I meant lim x->3 without the "+"

jbunniii's hint doesn't really depend on the limit being one sided. Think about it.

 

[Torshi]

jbunniii's hint doesn't really depend on the limit being one sided. Think about it.

can i just pick any number or? because any same number divided by itself = 1 unless the answer wouldn't fit with the other functions

 

[Dick]

can i just pick any number or? because any same number divided by itself = 1 unless the answer wouldn't fit with the other functions

You don't pick a number. You want to pick two functions of x. jbunniii's hint is that if lim x->3 of f(x) and g(x) both exist, you won't succeed. What an example of a function whose limit doesn't exist as x->3?

 

[Dick]

Is this part of the same problem as https://www.physicsforums.com/showthread.php?t=667668 ?

 

[Torshi]

You don't pick a number. You want to pick two functions of x. jbunniii's hint is that if lim x->3 of f(x) and g(x) both exist, you won't succeed. What an example of a function whose limit doesn't exist as x->3?

lim x-> 3 (x^2-9)/(x-3)? Sorry if I'm way off or all over the place? I'm unsure.

Edit:

(x-3)/(x-3) from the first part.

 

[Torshi]

Is this part of the same problem as https://www.physicsforums.com/showthread.php?t=667668 ?

That's the first part yes - is this needed? My fault if so.

 

[jbunniii]

You won't be able to use exactly the functions from the first part, but putting (x-3) in the denominator is a good idea. See if you can find f and g such that \lim_{x \rightarrow 3} f(x) = \infty and \lim_{x \rightarrow 3} g(x) = \infty which satisfy \lim_{x \rightarrow 3} \frac{f(x)}{g(x)} = 1 but \lim_{x \rightarrow 3} f(x) - g(x) \neq 0. (By choosing correctly, you can make that last limit equal to anything you want.)

 

[Dick]

That's the first part yes - is this needed? My fault if so.

No, no fault here. But they are interrelated enough that you should be able to use one to help solve the other. What about trying to use f(x)=x/(x-3) and g(x)=3/(x-3) on this problem as well? What are the limits of f(x)/g(x) and f(x)-g(x) as x->3?

 

[Torshi]

I hate to say it. But, it's not clicking for me. So far I have x-3 in the denominator.

I'm not too savvy with these functions and limits and am having trouble understanding the relations. I think I'm overlooking it most likely. These problems are more than we do in class purposely, so in class we don't do problems like these, so it's harder for me to comprehend what's going on when no examples like these have been taught. My apologies. It's quite frustrating.

 

[Dick]

I hate to say it. But, it's not clicking for me. So far I have x-3 in the denominator.

I'm not too savvy with these functions and limits and am having trouble understanding the relations. I think I'm overlooking it most likely. These problems are more than we do in class purposely, so in class we don't do problems like these, so it's harder for me to comprehend what's going on when no examples like these have been taught. My apologies. It's quite frustrating.

Hang in there, but didn't you already figure out lim x->3 of f(x)-g(x)=1 in the previous problem? That's half the battle. Remind me you how did it. Then use algebra to simplify f(x)/g(x) and try to find the limit of that. f(x)=x/(x-3) and g(x)=3/(x-3).

 

[Torshi]

Hang in there, but didn't you already figure out lim x->3 of f(x)-g(x)=1 in the previous problem? That's half the battle. Remind me how did it. Then use algebra to simplify f(x)/g(x) and try to find the limit of that. f(x)=x/(x-3) and g(x)=3/(x-3).

I honestly had a friend do the f(x)-g(x) = 1

But, I do understand from the first part that lim x->3+ (x/x-3) = 3/0+ = +∞
and for g(x) (3)/(x-3) = 3/0+ = +∞
but as for coming up with lim x->3+ (x-3)/(x-3) = 1 as a pair function with lim x->3+ (f(x)-g(x)) ≠0 I'm not sure.

 

[Dick]

I honestly had a friend do the f(x)-g(x) = 1

But, I do understand from the first part that lim x->3+ (x/x-3) = 3/0+ = +∞
and for g(x) (3)/(x-3) = 3/0+ = +∞
but as for coming up with lim x->3+ (x-3)/(x-3) = 1 as a pair function with lim x->3+ (f(x)-g(x)) ≠0 I'm not sure.

Well, that's honest. Is your algebra the problem? x/(x-3)-3/(x-3) should be easy to simplify. It's just like 1/3-2/3. Likewise, (x/(x-3))/(3/(x-3)) shouldn't be much of a challenge. It's just like (1/3)/(2/3). Can you do the number problems? Simplify the algebra problems with a variable the same way.

 

[Torshi]

Is your algebra the problem? x/(x-3)-3/(x-3) should be easy to simplify. It's just like 1/3-2/3. Likewise, (x/(x-3))/(3/(x-3)) shouldn't be much of a challenge. It's just like (1/3)/(2/3). Can you do the number problems? Simplify the algebra problems with a variable the same way.

isn't dividing fractions like the same as multiplying the inverse? So 1/3 / 2/3 = 3/6 = 1/2

My algebra isn't bad per se, it's just my direction in math is bad as of now. I don't know where to go when done with the question. With that being said I haven't added and subtracted fractions with variables in a while - I just looked it up now as a refresher, not bad at all! I remember now.

edit: I get how my friend got the answer now though! because the "-" carries over to the numerator which is 3 hence top and bottom are x-3 and I'm assuming two of the same variables equal out to 1

 

[Dick]

isn't dividing fractions like the same as multiplying the inverse? So 1/3 / 2/3 = 3/6 = 1/2?

My algebra isn't bad per se, it's just my direction in math is bad as of now. I don't know where to go when done with the question. With that being said I haven't added and subtracted fractions with variables in a while - I just looked it up now as a refresher, not bad at all! I remember now.

Yes, not so hard once you think back. So what are the simplified forms of f(x)-g(x) and f(x)/g(x)?? The first step is pure algebra. You don't think about the limits until later.

 

[Torshi]

As for lim x->3 f(x)/g(x)

(x)/(x-3) / (3)/(x-3)

(x)/(x-3) * (x-3)/(3)

= x/3
x=3 ? if this means anything since f(x)/g(x) = 1 ?

sorry I posted this before seeing your response.

 

[Dick]

As for lim x->3 f(x)/g(x)

(x)/(x-3) / (3)/(x-3)

(x)/(x-3) * (x-3)/(3)

= x/3
x=3 ? if this means anything since f(x)/g(x) = 1 ?

f(x)/g(x)=x/3, when x is not 3, sure. You've got the algebra part. Now you just have to figure out the limit x->3 x/3. You are writing '?' marks, but I'm not sure what the question is.

 

[Torshi]

f(x)/g(x)=x/3, when x is not 3, sure. You've got the algebra part. Now you just have to figure out the limit x->3 x/3. You are writing '?' marks, but I'm not sure what the question is.

Sorry my "?" is in regards to see if I'm on the right track.

As the lim x->3 x/3 = 1 since I just plugged in 3 so 3/3 = 1

 

[Dick]

Sorry my "?" is in regards to see if I'm on the right track.

As the lim x->3 x/3 = 1 since I just plugged in 3 so 3/3 = 1

Yes, you are on the right track. Ok, so you now know lim x->3 f(x)/g(x) is 1 and lim x->3 f(x)-g(x)=1. So isn't that a counterexample to the result they stated? Hence the statement is NOT true?

 

[Torshi]

Yes, you are on the right track. Ok, so you now know lim x->3 f(x)/g(x) is 1 and lim x-> f(x)-g(x)=1. So isn't that a counterexample to the result they stated?

Is it because both essentially equal 1? I do believe it is a counterexample to prove the statement to be false

Thank you! I appreciate it. This really helped me UNDERSTAND.

 

[Dick]

Is it because both essentially equal 1? I do believe it is a counterexample to prove the statement to be false.

Yes, it is. Not so bad, yes? But realizing that lim x->3 x/3=1 is only sort of like plugging in. What you should be thinking is that "if x gets closer and closer to 3 then x/3 gets closer and closer to 1". It's a fine distinction. With the original expressions like (x-3)/(x-3) you can't plug in. You need to use algebra to reduce them to expressions you can 'plug in'.

 

[Torshi]

Yes, it is. Not so bad, yes? But realizing that lim x->3 x/3=1 is only sort of like plugging in. What you should be thinking is that "if x gets closer and closer to 3 then x/3 gets closer and closer to 1". It's a fine distinction. With the original expressions like (x-3)/(x-3) you can't plug in. You need to use algebra to reduce them to expressions you can 'plug in'.

Well I do know for some problems you have to factor out and cancel out (not sure if correct term in these cases) top and bottom variables per se in order to get a value. Again thanks!

 

[Dick]

Well I do know for some problems you have to factor out and cancel out (not sure if correct term in these cases) top and bottom variables per se in order to get a value. Again thanks!

Yes, and these are like those. Very welcome!