Spivak Calculus Chapter 5 Limits Problem 22?
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Hmm, 4th edition, did I find another error in this book? (This turns out to be misunderstanding one word that makes a huge difference, edited)
It's Chapter 5 Question 22 about limits. The question and it's answer exactly is:
Question:
Consider a function $f$ with the following property: if $g$ is any function for which $lim_xto 0g(x)$ does not exist, then $lim_xto 0[f(x)+g(x)]$ also does not exist. Prove that this happens if and only if $lim_xto 0f(x)$ does exist. Hint: This is actually very easy: the assumption that $lim_xto 0f(x)$ does not exist leads to an immediate contradiction if you consider the right $g$.
Answer from Answer book:
If $lim_xto 0f(x)$ does exist, then it is clear that $lim_xto 0[f(x)+g(x)]$ does not exist whenever $lim_xto 0g(x)$ does not exist [this was Problem 8(b) and (c)]. On the other hand, if $lim_xto 0f(x)$ does not exist, choose $g=-f$; then $lim_xto 0g(x)$ does not exist, but $lim_xto 0[f(x)+g(x)]$ does exist.
I think the question is wrong or a typo(on "if and only if")? if $f(x)=1/x$ and $g(x)=1/x+1$, then $lim_xto 0[f(x)+g(x)]$ does not exist.
And the answer from "On the other hand" then on is point less, because randomly choose $g=-f$ only proves something can be true/false, but not must be true/false.
limits
asked Sep 9 at 9:40
user401653
406
add a comment |Â
up vote
3
down vote
favorite
Hmm, 4th edition, did I find another error in this book? (This turns out to be misunderstanding one word that makes a huge difference, edited)
It's Chapter 5 Question 22 about limits. The question and it's answer exactly is:
Question:
Consider a function $f$ with the following property: if $g$ is any function for which $lim_xto 0g(x)$ does not exist, then $lim_xto 0[f(x)+g(x)]$ also does not exist. Prove that this happens if and only if $lim_xto 0f(x)$ does exist. Hint: This is actually very easy: the assumption that $lim_xto 0f(x)$ does not exist leads to an immediate contradiction if you consider the right $g$.
Answer from Answer book:
If $lim_xto 0f(x)$ does exist, then it is clear that $lim_xto 0[f(x)+g(x)]$ does not exist whenever $lim_xto 0g(x)$ does not exist [this was Problem 8(b) and (c)]. On the other hand, if $lim_xto 0f(x)$ does not exist, choose $g=-f$; then $lim_xto 0g(x)$ does not exist, but $lim_xto 0[f(x)+g(x)]$ does exist.
I think the question is wrong or a typo(on "if and only if")? if $f(x)=1/x$ and $g(x)=1/x+1$, then $lim_xto 0[f(x)+g(x)]$ does not exist.
And the answer from "On the other hand" then on is point less, because randomly choose $g=-f$ only proves something can be true/false, but not must be true/false.
limits
asked Sep 9 at 9:40
user401653
406
Your reasoning is sound. I also think you understood what the exercise meant to convey. I also found this related discussion: math.stackexchange.com/questions/562938/…
– Imago
Sep 9 at 9:47
Can we make the title perhaps less mysterious and include information about the supposed error, rather than the location of it?
– Asaf Karagila♦
Sep 9 at 11:00
Mathematicians should stop using the word "any".
– yo'
Sep 9 at 15:45
Edited the title for other's quick search with the same problem.
– user401653
Sep 10 at 0:52
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Hmm, 4th edition, did I find another error in this book? (This turns out to be misunderstanding one word that makes a huge difference, edited)
It's Chapter 5 Question 22 about limits. The question and it's answer exactly is:
Question:
Consider a function $f$ with the following property: if $g$ is any function for which $lim_xto 0g(x)$ does not exist, then $lim_xto 0[f(x)+g(x)]$ also does not exist. Prove that this happens if and only if $lim_xto 0f(x)$ does exist. Hint: This is actually very easy: the assumption that $lim_xto 0f(x)$ does not exist leads to an immediate contradiction if you consider the right $g$.
Answer from Answer book:
If $lim_xto 0f(x)$ does exist, then it is clear that $lim_xto 0[f(x)+g(x)]$ does not exist whenever $lim_xto 0g(x)$ does not exist [this was Problem 8(b) and (c)]. On the other hand, if $lim_xto 0f(x)$ does not exist, choose $g=-f$; then $lim_xto 0g(x)$ does not exist, but $lim_xto 0[f(x)+g(x)]$ does exist.
I think the question is wrong or a typo(on "if and only if")? if $f(x)=1/x$ and $g(x)=1/x+1$, then $lim_xto 0[f(x)+g(x)]$ does not exist.
And the answer from "On the other hand" then on is point less, because randomly choose $g=-f$ only proves something can be true/false, but not must be true/false.
limits
asked Sep 9 at 9:40
user401653
406
Hmm, 4th edition, did I find another error in this book? (This turns out to be misunderstanding one word that makes a huge difference, edited)
It's Chapter 5 Question 22 about limits. The question and it's answer exactly is:
Question:
Consider a function $f$ with the following property: if $g$ is any function for which $lim_xto 0g(x)$ does not exist, then $lim_xto 0[f(x)+g(x)]$ also does not exist. Prove that this happens if and only if $lim_xto 0f(x)$ does exist. Hint: This is actually very easy: the assumption that $lim_xto 0f(x)$ does not exist leads to an immediate contradiction if you consider the right $g$.
Answer from Answer book:
If $lim_xto 0f(x)$ does exist, then it is clear that $lim_xto 0[f(x)+g(x)]$ does not exist whenever $lim_xto 0g(x)$ does not exist [this was Problem 8(b) and (c)]. On the other hand, if $lim_xto 0f(x)$ does not exist, choose $g=-f$; then $lim_xto 0g(x)$ does not exist, but $lim_xto 0[f(x)+g(x)]$ does exist.
I think the question is wrong or a typo(on "if and only if")? if $f(x)=1/x$ and $g(x)=1/x+1$, then $lim_xto 0[f(x)+g(x)]$ does not exist.
And the answer from "On the other hand" then on is point less, because randomly choose $g=-f$ only proves something can be true/false, but not must be true/false.
limits
limits
asked Sep 9 at 9:40
user401653
406
asked Sep 9 at 9:40
user401653
406
edited Sep 10 at 0:48
asked Sep 9 at 9:40
user401653
406
asked Sep 9 at 9:40
user401653
406
asked Sep 9 at 9:40
user401653
406
406
Your reasoning is sound. I also think you understood what the exercise meant to convey. I also found this related discussion: math.stackexchange.com/questions/562938/…
– Imago
Sep 9 at 9:47
Can we make the title perhaps less mysterious and include information about the supposed error, rather than the location of it?
– Asaf Karagila♦
Sep 9 at 11:00
Mathematicians should stop using the word "any".
– yo'
Sep 9 at 15:45
Edited the title for other's quick search with the same problem.
– user401653
Sep 10 at 0:52
add a comment |Â
Your reasoning is sound. I also think you understood what the exercise meant to convey. I also found this related discussion: math.stackexchange.com/questions/562938/…
– Imago
Sep 9 at 9:47
Can we make the title perhaps less mysterious and include information about the supposed error, rather than the location of it?
– Asaf Karagila♦
Sep 9 at 11:00
Mathematicians should stop using the word "any".
– yo'
Sep 9 at 15:45
Edited the title for other's quick search with the same problem.
– user401653
Sep 10 at 0:52
Your reasoning is sound. I also think you understood what the exercise meant to convey. I also found this related discussion: math.stackexchange.com/questions/562938/…
– Imago
Sep 9 at 9:47
Your reasoning is sound. I also think you understood what the exercise meant to convey. I also found this related discussion: math.stackexchange.com/questions/562938/…
– Imago
Sep 9 at 9:47
Can we make the title perhaps less mysterious and include information about the supposed error, rather than the location of it?
– Asaf Karagila♦
Sep 9 at 11:00
Can we make the title perhaps less mysterious and include information about the supposed error, rather than the location of it?
– Asaf Karagila♦
Sep 9 at 11:00
Mathematicians should stop using the word "any".
– yo'
Sep 9 at 15:45
Mathematicians should stop using the word "any".
– yo'
Sep 9 at 15:45
Edited the title for other's quick search with the same problem.
– user401653
Sep 10 at 0:52
Edited the title for other's quick search with the same problem.
– user401653
Sep 10 at 0:52
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
8
down vote
accepted
You may have misunderstood the question. The statement is, slightly reformulated to stress the point you may have gotten wrong
if you are given $f$, such that for every $g$ with $lim_xrightarrow 0 g(x)$ not existing you can conlude that $f+g$ does not have a limit at $x=0$, then this is equivalent to $lim_xrightarrow 0 f(x)$ exists.
The point you may have ignored that you may choose $g$ arbitrarily, after $f$ is given. The proof is the same you copied in your question.
I'm not sure what your counter example is supposed to prove. The point is that you may in fact choose any $g$, this is the other direction in the proof.
To make it more explicit: If $f(x) = frac1x$ then $lim_xrightarrow 0 (f(x) + (-frac1x))=0$, which means there is a function $g$ such that $lim_xrightarrow 0 g(x)$ does not exist (namely $-f$) while $f+g$ does has a limit at $0$, hence $f$ need not satisfy the conclusion (i.e. the statement is not applicable).
answered Sep 9 at 10:03
Thomas
16.1k21430
Thank you Thomas! I think I start to understand a bit of your answer. If the $g$ is not a particular one, but every or any, then the question maybe more complicate than I thought, I must go jogging and read your answer later.
– user401653
Sep 9 at 10:51
Add an exist limit to a non-exist limit will not make it exist. A non-exist limit must added to a non-exist limit to make it exist, and there will always be at least one such function to make it happen. I think this is the problem is telling about.
– user401653
Sep 10 at 1:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
You may have misunderstood the question. The statement is, slightly reformulated to stress the point you may have gotten wrong
if you are given $f$, such that for every $g$ with $lim_xrightarrow 0 g(x)$ not existing you can conlude that $f+g$ does not have a limit at $x=0$, then this is equivalent to $lim_xrightarrow 0 f(x)$ exists.
The point you may have ignored that you may choose $g$ arbitrarily, after $f$ is given. The proof is the same you copied in your question.
I'm not sure what your counter example is supposed to prove. The point is that you may in fact choose any $g$, this is the other direction in the proof.
To make it more explicit: If $f(x) = frac1x$ then $lim_xrightarrow 0 (f(x) + (-frac1x))=0$, which means there is a function $g$ such that $lim_xrightarrow 0 g(x)$ does not exist (namely $-f$) while $f+g$ does has a limit at $0$, hence $f$ need not satisfy the conclusion (i.e. the statement is not applicable).
answered Sep 9 at 10:03
Thomas
16.1k21430
Thank you Thomas! I think I start to understand a bit of your answer. If the $g$ is not a particular one, but every or any, then the question maybe more complicate than I thought, I must go jogging and read your answer later.
– user401653
Sep 9 at 10:51
Add an exist limit to a non-exist limit will not make it exist. A non-exist limit must added to a non-exist limit to make it exist, and there will always be at least one such function to make it happen. I think this is the problem is telling about.
– user401653
Sep 10 at 1:07
add a comment |Â
up vote
8
down vote
accepted
You may have misunderstood the question. The statement is, slightly reformulated to stress the point you may have gotten wrong
if you are given $f$, such that for every $g$ with $lim_xrightarrow 0 g(x)$ not existing you can conlude that $f+g$ does not have a limit at $x=0$, then this is equivalent to $lim_xrightarrow 0 f(x)$ exists.
The point you may have ignored that you may choose $g$ arbitrarily, after $f$ is given. The proof is the same you copied in your question.
I'm not sure what your counter example is supposed to prove. The point is that you may in fact choose any $g$, this is the other direction in the proof.
To make it more explicit: If $f(x) = frac1x$ then $lim_xrightarrow 0 (f(x) + (-frac1x))=0$, which means there is a function $g$ such that $lim_xrightarrow 0 g(x)$ does not exist (namely $-f$) while $f+g$ does has a limit at $0$, hence $f$ need not satisfy the conclusion (i.e. the statement is not applicable).
answered Sep 9 at 10:03
Thomas
16.1k21430
Thank you Thomas! I think I start to understand a bit of your answer. If the $g$ is not a particular one, but every or any, then the question maybe more complicate than I thought, I must go jogging and read your answer later.
– user401653
Sep 9 at 10:51
Add an exist limit to a non-exist limit will not make it exist. A non-exist limit must added to a non-exist limit to make it exist, and there will always be at least one such function to make it happen. I think this is the problem is telling about.
– user401653
Sep 10 at 1:07
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
You may have misunderstood the question. The statement is, slightly reformulated to stress the point you may have gotten wrong
if you are given $f$, such that for every $g$ with $lim_xrightarrow 0 g(x)$ not existing you can conlude that $f+g$ does not have a limit at $x=0$, then this is equivalent to $lim_xrightarrow 0 f(x)$ exists.
The point you may have ignored that you may choose $g$ arbitrarily, after $f$ is given. The proof is the same you copied in your question.
I'm not sure what your counter example is supposed to prove. The point is that you may in fact choose any $g$, this is the other direction in the proof.
To make it more explicit: If $f(x) = frac1x$ then $lim_xrightarrow 0 (f(x) + (-frac1x))=0$, which means there is a function $g$ such that $lim_xrightarrow 0 g(x)$ does not exist (namely $-f$) while $f+g$ does has a limit at $0$, hence $f$ need not satisfy the conclusion (i.e. the statement is not applicable).
answered Sep 9 at 10:03
Thomas
16.1k21430
You may have misunderstood the question. The statement is, slightly reformulated to stress the point you may have gotten wrong
if you are given $f$, such that for every $g$ with $lim_xrightarrow 0 g(x)$ not existing you can conlude that $f+g$ does not have a limit at $x=0$, then this is equivalent to $lim_xrightarrow 0 f(x)$ exists.
The point you may have ignored that you may choose $g$ arbitrarily, after $f$ is given. The proof is the same you copied in your question.
I'm not sure what your counter example is supposed to prove. The point is that you may in fact choose any $g$, this is the other direction in the proof.
To make it more explicit: If $f(x) = frac1x$ then $lim_xrightarrow 0 (f(x) + (-frac1x))=0$, which means there is a function $g$ such that $lim_xrightarrow 0 g(x)$ does not exist (namely $-f$) while $f+g$ does has a limit at $0$, hence $f$ need not satisfy the conclusion (i.e. the statement is not applicable).
answered Sep 9 at 10:03
Thomas
16.1k21430
edited Sep 9 at 10:27
answered Sep 9 at 10:03
Thomas
16.1k21430
answered Sep 9 at 10:03
Thomas
16.1k21430
answered Sep 9 at 10:03
Thomas
16.1k21430
16.1k21430
Thank you Thomas! I think I start to understand a bit of your answer. If the $g$ is not a particular one, but every or any, then the question maybe more complicate than I thought, I must go jogging and read your answer later.
– user401653
Sep 9 at 10:51
Add an exist limit to a non-exist limit will not make it exist. A non-exist limit must added to a non-exist limit to make it exist, and there will always be at least one such function to make it happen. I think this is the problem is telling about.
– user401653
Sep 10 at 1:07
add a comment |Â
Thank you Thomas! I think I start to understand a bit of your answer. If the $g$ is not a particular one, but every or any, then the question maybe more complicate than I thought, I must go jogging and read your answer later.
– user401653
Sep 9 at 10:51
Add an exist limit to a non-exist limit will not make it exist. A non-exist limit must added to a non-exist limit to make it exist, and there will always be at least one such function to make it happen. I think this is the problem is telling about.
– user401653
Sep 10 at 1:07
Thank you Thomas! I think I start to understand a bit of your answer. If the $g$ is not a particular one, but every or any, then the question maybe more complicate than I thought, I must go jogging and read your answer later.
– user401653
Sep 9 at 10:51
Thank you Thomas! I think I start to understand a bit of your answer. If the $g$ is not a particular one, but every or any, then the question maybe more complicate than I thought, I must go jogging and read your answer later.
– user401653
Sep 9 at 10:51
Add an exist limit to a non-exist limit will not make it exist. A non-exist limit must added to a non-exist limit to make it exist, and there will always be at least one such function to make it happen. I think this is the problem is telling about.
– user401653
Sep 10 at 1:07
Add an exist limit to a non-exist limit will not make it exist. A non-exist limit must added to a non-exist limit to make it exist, and there will always be at least one such function to make it happen. I think this is the problem is telling about.
– user401653
Sep 10 at 1:07
add a comment |Â
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Your reasoning is sound. I also think you understood what the exercise meant to convey. I also found this related discussion: math.stackexchange.com/questions/562938/…
– Imago
Sep 9 at 9:47
Can we make the title perhaps less mysterious and include information about the supposed error, rather than the location of it?
– Asaf Karagila♦
Sep 9 at 11:00
Mathematicians should stop using the word "any".
– yo'
Sep 9 at 15:45
Edited the title for other's quick search with the same problem.
– user401653
Sep 10 at 0:52