Spivak, Differentiability (An Exercise)
I am working on a problem from Spivak, question 52b. I will type part a for reference as well since it may be relevant to the problem at hand. It is stated as follows:
There is another form of L'Hôpital's Rule which requires more than algebraic
manipulations: If $lim_{xrightarrow infty}f(x)=lim_{xrightarrow infty}g(x)=infty$ and $lim_{xrightarrow infty}frac{f'(x)}{g'(x)}=l$ then $lim_{xrightarrow infty}frac{f(x)}{g(x)}=l$. Prove this as follows.
(a) For every $varepsilon>0$ there is a number $a$ such that $$left|frac{f'(x)}{g'(x)}-lright| <varepsilonqquadtext{for }x>a.$$
Apply the Cauchy Mean Value Theorem to $f$ and $g$ on $[a,x]$ to show that $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<varepsilonqquadtext{for }x>a.$$
(Why can we assume that $g(x)-g(a) not=0?$)
(b) Now write
$$frac{f(x)}{g(x)}=frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}$$
(why can we assume that $f(x)-f(a)not=0$ for large x?) and conclude that
$$left|frac{f(x)}{g(x)}-lright|<2varepsilonquadtext{for sufficiently large }x.$$
For part b, which is where my troubles lie, I have dealt with the limits $lim_{xrightarrow infty}frac{f(x)}{f(x)-f(a)}$ and $lim_{xrightarrow infty}frac{g(x)-g(a)}{g(x)}$ and I was able to conclude both are 1, and the limit $lim_{xrightarrow infty}frac{f(x)-f(a)}{g(x)-g(a)}$ was dealt with in part a via Cauchy MVT. And we know that for large enough $x$ (say, $x>N$), $f$ is unbounded so we can infer that $f(x)-f(a) not=0$ for large enough $x$. Now this is all fine, but I cannot for the life of me figure out where a $2epsilon$ comes from. I have been playing with this for a couple days with no luck.
Thanks ahead of time for any help.
calculus
edited Dec 1 at 10:33
Luiz Cordeiro
12.5k1143
asked Dec 1 at 9:59
kyle campbell
244
add a comment |
I am working on a problem from Spivak, question 52b. I will type part a for reference as well since it may be relevant to the problem at hand. It is stated as follows:
There is another form of L'Hôpital's Rule which requires more than algebraic
manipulations: If $lim_{xrightarrow infty}f(x)=lim_{xrightarrow infty}g(x)=infty$ and $lim_{xrightarrow infty}frac{f'(x)}{g'(x)}=l$ then $lim_{xrightarrow infty}frac{f(x)}{g(x)}=l$. Prove this as follows.
(a) For every $varepsilon>0$ there is a number $a$ such that $$left|frac{f'(x)}{g'(x)}-lright| <varepsilonqquadtext{for }x>a.$$
Apply the Cauchy Mean Value Theorem to $f$ and $g$ on $[a,x]$ to show that $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<varepsilonqquadtext{for }x>a.$$
(Why can we assume that $g(x)-g(a) not=0?$)
(b) Now write
$$frac{f(x)}{g(x)}=frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}$$
(why can we assume that $f(x)-f(a)not=0$ for large x?) and conclude that
$$left|frac{f(x)}{g(x)}-lright|<2varepsilonquadtext{for sufficiently large }x.$$
For part b, which is where my troubles lie, I have dealt with the limits $lim_{xrightarrow infty}frac{f(x)}{f(x)-f(a)}$ and $lim_{xrightarrow infty}frac{g(x)-g(a)}{g(x)}$ and I was able to conclude both are 1, and the limit $lim_{xrightarrow infty}frac{f(x)-f(a)}{g(x)-g(a)}$ was dealt with in part a via Cauchy MVT. And we know that for large enough $x$ (say, $x>N$), $f$ is unbounded so we can infer that $f(x)-f(a) not=0$ for large enough $x$. Now this is all fine, but I cannot for the life of me figure out where a $2epsilon$ comes from. I have been playing with this for a couple days with no luck.
Thanks ahead of time for any help.
calculus
edited Dec 1 at 10:33
Luiz Cordeiro
12.5k1143
asked Dec 1 at 9:59
kyle campbell
244
Well note that by assumption $g'(x) $ is non-zero after a certain value of $x$ say $a$ and by Darboux theorem it is of constant sign. Thus $g$ is strictly monotone in $[a, infty) $ and that means $g(x) neq g(a)$ if $x>a$. The other part is mostly the typical $epsilon, delta$ gymnastics which involves triangle inequality and some simple bounds for various terms.
– Paramanand Singh
Dec 1 at 15:50
add a comment |
I am working on a problem from Spivak, question 52b. I will type part a for reference as well since it may be relevant to the problem at hand. It is stated as follows:
There is another form of L'Hôpital's Rule which requires more than algebraic
manipulations: If $lim_{xrightarrow infty}f(x)=lim_{xrightarrow infty}g(x)=infty$ and $lim_{xrightarrow infty}frac{f'(x)}{g'(x)}=l$ then $lim_{xrightarrow infty}frac{f(x)}{g(x)}=l$. Prove this as follows.
(a) For every $varepsilon>0$ there is a number $a$ such that $$left|frac{f'(x)}{g'(x)}-lright| <varepsilonqquadtext{for }x>a.$$
Apply the Cauchy Mean Value Theorem to $f$ and $g$ on $[a,x]$ to show that $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<varepsilonqquadtext{for }x>a.$$
(Why can we assume that $g(x)-g(a) not=0?$)
(b) Now write
$$frac{f(x)}{g(x)}=frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}$$
(why can we assume that $f(x)-f(a)not=0$ for large x?) and conclude that
$$left|frac{f(x)}{g(x)}-lright|<2varepsilonquadtext{for sufficiently large }x.$$
For part b, which is where my troubles lie, I have dealt with the limits $lim_{xrightarrow infty}frac{f(x)}{f(x)-f(a)}$ and $lim_{xrightarrow infty}frac{g(x)-g(a)}{g(x)}$ and I was able to conclude both are 1, and the limit $lim_{xrightarrow infty}frac{f(x)-f(a)}{g(x)-g(a)}$ was dealt with in part a via Cauchy MVT. And we know that for large enough $x$ (say, $x>N$), $f$ is unbounded so we can infer that $f(x)-f(a) not=0$ for large enough $x$. Now this is all fine, but I cannot for the life of me figure out where a $2epsilon$ comes from. I have been playing with this for a couple days with no luck.
Thanks ahead of time for any help.
calculus
edited Dec 1 at 10:33
Luiz Cordeiro
12.5k1143
asked Dec 1 at 9:59
kyle campbell
244
I am working on a problem from Spivak, question 52b. I will type part a for reference as well since it may be relevant to the problem at hand. It is stated as follows:
There is another form of L'Hôpital's Rule which requires more than algebraic
manipulations: If $lim_{xrightarrow infty}f(x)=lim_{xrightarrow infty}g(x)=infty$ and $lim_{xrightarrow infty}frac{f'(x)}{g'(x)}=l$ then $lim_{xrightarrow infty}frac{f(x)}{g(x)}=l$. Prove this as follows.
(a) For every $varepsilon>0$ there is a number $a$ such that $$left|frac{f'(x)}{g'(x)}-lright| <varepsilonqquadtext{for }x>a.$$
Apply the Cauchy Mean Value Theorem to $f$ and $g$ on $[a,x]$ to show that $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<varepsilonqquadtext{for }x>a.$$
(Why can we assume that $g(x)-g(a) not=0?$)
(b) Now write
$$frac{f(x)}{g(x)}=frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}$$
(why can we assume that $f(x)-f(a)not=0$ for large x?) and conclude that
$$left|frac{f(x)}{g(x)}-lright|<2varepsilonquadtext{for sufficiently large }x.$$
For part b, which is where my troubles lie, I have dealt with the limits $lim_{xrightarrow infty}frac{f(x)}{f(x)-f(a)}$ and $lim_{xrightarrow infty}frac{g(x)-g(a)}{g(x)}$ and I was able to conclude both are 1, and the limit $lim_{xrightarrow infty}frac{f(x)-f(a)}{g(x)-g(a)}$ was dealt with in part a via Cauchy MVT. And we know that for large enough $x$ (say, $x>N$), $f$ is unbounded so we can infer that $f(x)-f(a) not=0$ for large enough $x$. Now this is all fine, but I cannot for the life of me figure out where a $2epsilon$ comes from. I have been playing with this for a couple days with no luck.
Thanks ahead of time for any help.
calculus
calculus
edited Dec 1 at 10:33
Luiz Cordeiro
12.5k1143
asked Dec 1 at 9:59
kyle campbell
244
edited Dec 1 at 10:33
Luiz Cordeiro
12.5k1143
asked Dec 1 at 9:59
kyle campbell
244
edited Dec 1 at 10:33
Luiz Cordeiro
12.5k1143
edited Dec 1 at 10:33
Luiz Cordeiro
12.5k1143
edited Dec 1 at 10:33
Luiz Cordeiro
12.5k1143
12.5k1143
asked Dec 1 at 9:59
kyle campbell
244
asked Dec 1 at 9:59
kyle campbell
244
asked Dec 1 at 9:59
kyle campbell
244
244
Well note that by assumption $g'(x) $ is non-zero after a certain value of $x$ say $a$ and by Darboux theorem it is of constant sign. Thus $g$ is strictly monotone in $[a, infty) $ and that means $g(x) neq g(a)$ if $x>a$. The other part is mostly the typical $epsilon, delta$ gymnastics which involves triangle inequality and some simple bounds for various terms.
– Paramanand Singh
Dec 1 at 15:50
add a comment |
Well note that by assumption $g'(x) $ is non-zero after a certain value of $x$ say $a$ and by Darboux theorem it is of constant sign. Thus $g$ is strictly monotone in $[a, infty) $ and that means $g(x) neq g(a)$ if $x>a$. The other part is mostly the typical $epsilon, delta$ gymnastics which involves triangle inequality and some simple bounds for various terms.
– Paramanand Singh
Dec 1 at 15:50
Well note that by assumption $g'(x) $ is non-zero after a certain value of $x$ say $a$ and by Darboux theorem it is of constant sign. Thus $g$ is strictly monotone in $[a, infty) $ and that means $g(x) neq g(a)$ if $x>a$. The other part is mostly the typical $epsilon, delta$ gymnastics which involves triangle inequality and some simple bounds for various terms.
– Paramanand Singh
Dec 1 at 15:50
Well note that by assumption $g'(x) $ is non-zero after a certain value of $x$ say $a$ and by Darboux theorem it is of constant sign. Thus $g$ is strictly monotone in $[a, infty) $ and that means $g(x) neq g(a)$ if $x>a$. The other part is mostly the typical $epsilon, delta$ gymnastics which involves triangle inequality and some simple bounds for various terms.
– Paramanand Singh
Dec 1 at 15:50
add a comment |
1 Answer
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Basically, write down whatevver the hint says. You will have an expression of the form $|A(x)-l|$ for some $A(x)$. Use the first part and write $|A(x)-l|leq |A(x)-frac{f(x)-f(a)}{g(x)-g(a)}|+|frac{f(x)-f(a)}{g(x)-g(a)}-l|$ and try to work from there. Below are the details.
begin{align*}
left|frac{f(x)}{g(x)}-lright|&=left|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-lright|\
&=left|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-frac{f(x)-f(a)}{g(x)-g(a)}+frac{f(x)-f(a)}{g(x)-g(a)}-lright|\
&leqleft|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-frac{f(x)-f(a)}{g(x)-g(a)}right|+left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|\
&=left|frac{f(x)-f(a)}{g(x)-g(a)}right|cdotleft|frac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-1right|+left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|
end{align*}
The term $left|frac{f(x)-f(a)}{g(x)-g(a)}right|$ gets close to $|l|$ (so it may be bounded by $|l|+1$), the term mutiplying it gets close to $0$, and the last term gets close to $0$ as well, so I'll leave the details to you
answered Dec 1 at 10:47
Luiz Cordeiro
12.5k1143
Thank you! Forgive me, since I must be missing something obvious, but I am still not seeing how a 2$varepsilon$ pops out. Any hints?
– kyle campbell
Dec 1 at 20:56
1
For large $x$, we have $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<1,$$ so $$left|frac{f(x)-f(a)}{g(x)-g(a)}right|leq |l|+1.$$ For $x$ sufficiently large, we also have $$left|frac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-1right|<varepsilon/2(|l|+1)$$ and $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<varepsilon/2$$ so substituting all these inequalities in the large equation in the answer above gives the term $2varepsilon$. The key is that item (a) holds for any $varepsilon$, so in particular for anything of the form $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:14
1
(cont.) How large $x$ needs to be will depend, however, on your given $varepsilon$, $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:17
add a comment |
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Basically, write down whatevver the hint says. You will have an expression of the form $|A(x)-l|$ for some $A(x)$. Use the first part and write $|A(x)-l|leq |A(x)-frac{f(x)-f(a)}{g(x)-g(a)}|+|frac{f(x)-f(a)}{g(x)-g(a)}-l|$ and try to work from there. Below are the details.
begin{align*}
left|frac{f(x)}{g(x)}-lright|&=left|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-lright|\
&=left|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-frac{f(x)-f(a)}{g(x)-g(a)}+frac{f(x)-f(a)}{g(x)-g(a)}-lright|\
&leqleft|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-frac{f(x)-f(a)}{g(x)-g(a)}right|+left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|\
&=left|frac{f(x)-f(a)}{g(x)-g(a)}right|cdotleft|frac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-1right|+left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|
end{align*}
The term $left|frac{f(x)-f(a)}{g(x)-g(a)}right|$ gets close to $|l|$ (so it may be bounded by $|l|+1$), the term mutiplying it gets close to $0$, and the last term gets close to $0$ as well, so I'll leave the details to you
answered Dec 1 at 10:47
Luiz Cordeiro
12.5k1143
Thank you! Forgive me, since I must be missing something obvious, but I am still not seeing how a 2$varepsilon$ pops out. Any hints?
– kyle campbell
Dec 1 at 20:56
1
For large $x$, we have $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<1,$$ so $$left|frac{f(x)-f(a)}{g(x)-g(a)}right|leq |l|+1.$$ For $x$ sufficiently large, we also have $$left|frac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-1right|<varepsilon/2(|l|+1)$$ and $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<varepsilon/2$$ so substituting all these inequalities in the large equation in the answer above gives the term $2varepsilon$. The key is that item (a) holds for any $varepsilon$, so in particular for anything of the form $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:14
1
(cont.) How large $x$ needs to be will depend, however, on your given $varepsilon$, $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:17
add a comment |
Basically, write down whatevver the hint says. You will have an expression of the form $|A(x)-l|$ for some $A(x)$. Use the first part and write $|A(x)-l|leq |A(x)-frac{f(x)-f(a)}{g(x)-g(a)}|+|frac{f(x)-f(a)}{g(x)-g(a)}-l|$ and try to work from there. Below are the details.
begin{align*}
left|frac{f(x)}{g(x)}-lright|&=left|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-lright|\
&=left|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-frac{f(x)-f(a)}{g(x)-g(a)}+frac{f(x)-f(a)}{g(x)-g(a)}-lright|\
&leqleft|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-frac{f(x)-f(a)}{g(x)-g(a)}right|+left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|\
&=left|frac{f(x)-f(a)}{g(x)-g(a)}right|cdotleft|frac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-1right|+left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|
end{align*}
The term $left|frac{f(x)-f(a)}{g(x)-g(a)}right|$ gets close to $|l|$ (so it may be bounded by $|l|+1$), the term mutiplying it gets close to $0$, and the last term gets close to $0$ as well, so I'll leave the details to you
answered Dec 1 at 10:47
Luiz Cordeiro
12.5k1143
Thank you! Forgive me, since I must be missing something obvious, but I am still not seeing how a 2$varepsilon$ pops out. Any hints?
– kyle campbell
Dec 1 at 20:56
1
For large $x$, we have $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<1,$$ so $$left|frac{f(x)-f(a)}{g(x)-g(a)}right|leq |l|+1.$$ For $x$ sufficiently large, we also have $$left|frac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-1right|<varepsilon/2(|l|+1)$$ and $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<varepsilon/2$$ so substituting all these inequalities in the large equation in the answer above gives the term $2varepsilon$. The key is that item (a) holds for any $varepsilon$, so in particular for anything of the form $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:14
1
(cont.) How large $x$ needs to be will depend, however, on your given $varepsilon$, $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:17
add a comment |
Basically, write down whatevver the hint says. You will have an expression of the form $|A(x)-l|$ for some $A(x)$. Use the first part and write $|A(x)-l|leq |A(x)-frac{f(x)-f(a)}{g(x)-g(a)}|+|frac{f(x)-f(a)}{g(x)-g(a)}-l|$ and try to work from there. Below are the details.
begin{align*}
left|frac{f(x)}{g(x)}-lright|&=left|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-lright|\
&=left|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-frac{f(x)-f(a)}{g(x)-g(a)}+frac{f(x)-f(a)}{g(x)-g(a)}-lright|\
&leqleft|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-frac{f(x)-f(a)}{g(x)-g(a)}right|+left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|\
&=left|frac{f(x)-f(a)}{g(x)-g(a)}right|cdotleft|frac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-1right|+left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|
end{align*}
The term $left|frac{f(x)-f(a)}{g(x)-g(a)}right|$ gets close to $|l|$ (so it may be bounded by $|l|+1$), the term mutiplying it gets close to $0$, and the last term gets close to $0$ as well, so I'll leave the details to you
answered Dec 1 at 10:47
Luiz Cordeiro
12.5k1143
Basically, write down whatevver the hint says. You will have an expression of the form $|A(x)-l|$ for some $A(x)$. Use the first part and write $|A(x)-l|leq |A(x)-frac{f(x)-f(a)}{g(x)-g(a)}|+|frac{f(x)-f(a)}{g(x)-g(a)}-l|$ and try to work from there. Below are the details.
begin{align*}
left|frac{f(x)}{g(x)}-lright|&=left|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-lright|\
&=left|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-frac{f(x)-f(a)}{g(x)-g(a)}+frac{f(x)-f(a)}{g(x)-g(a)}-lright|\
&leqleft|frac{f(x)-f(a)}{g(x)-g(a)}cdotfrac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-frac{f(x)-f(a)}{g(x)-g(a)}right|+left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|\
&=left|frac{f(x)-f(a)}{g(x)-g(a)}right|cdotleft|frac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-1right|+left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|
end{align*}
The term $left|frac{f(x)-f(a)}{g(x)-g(a)}right|$ gets close to $|l|$ (so it may be bounded by $|l|+1$), the term mutiplying it gets close to $0$, and the last term gets close to $0$ as well, so I'll leave the details to you
answered Dec 1 at 10:47
Luiz Cordeiro
12.5k1143
answered Dec 1 at 10:47
Luiz Cordeiro
12.5k1143
answered Dec 1 at 10:47
Luiz Cordeiro
12.5k1143
answered Dec 1 at 10:47
Luiz Cordeiro
12.5k1143
12.5k1143
Thank you! Forgive me, since I must be missing something obvious, but I am still not seeing how a 2$varepsilon$ pops out. Any hints?
– kyle campbell
Dec 1 at 20:56
1
For large $x$, we have $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<1,$$ so $$left|frac{f(x)-f(a)}{g(x)-g(a)}right|leq |l|+1.$$ For $x$ sufficiently large, we also have $$left|frac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-1right|<varepsilon/2(|l|+1)$$ and $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<varepsilon/2$$ so substituting all these inequalities in the large equation in the answer above gives the term $2varepsilon$. The key is that item (a) holds for any $varepsilon$, so in particular for anything of the form $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:14
1
(cont.) How large $x$ needs to be will depend, however, on your given $varepsilon$, $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:17
add a comment |
Thank you! Forgive me, since I must be missing something obvious, but I am still not seeing how a 2$varepsilon$ pops out. Any hints?
– kyle campbell
Dec 1 at 20:56
1
For large $x$, we have $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<1,$$ so $$left|frac{f(x)-f(a)}{g(x)-g(a)}right|leq |l|+1.$$ For $x$ sufficiently large, we also have $$left|frac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-1right|<varepsilon/2(|l|+1)$$ and $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<varepsilon/2$$ so substituting all these inequalities in the large equation in the answer above gives the term $2varepsilon$. The key is that item (a) holds for any $varepsilon$, so in particular for anything of the form $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:14
1
(cont.) How large $x$ needs to be will depend, however, on your given $varepsilon$, $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:17
Thank you! Forgive me, since I must be missing something obvious, but I am still not seeing how a 2$varepsilon$ pops out. Any hints?
– kyle campbell
Dec 1 at 20:56
Thank you! Forgive me, since I must be missing something obvious, but I am still not seeing how a 2$varepsilon$ pops out. Any hints?
– kyle campbell
Dec 1 at 20:56
1
1
For large $x$, we have $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<1,$$ so $$left|frac{f(x)-f(a)}{g(x)-g(a)}right|leq |l|+1.$$ For $x$ sufficiently large, we also have $$left|frac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-1right|<varepsilon/2(|l|+1)$$ and $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<varepsilon/2$$ so substituting all these inequalities in the large equation in the answer above gives the term $2varepsilon$. The key is that item (a) holds for any $varepsilon$, so in particular for anything of the form $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:14
For large $x$, we have $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<1,$$ so $$left|frac{f(x)-f(a)}{g(x)-g(a)}right|leq |l|+1.$$ For $x$ sufficiently large, we also have $$left|frac{f(x)}{f(x)-f(a)}cdotfrac{g(x)-g(a)}{g(x)}-1right|<varepsilon/2(|l|+1)$$ and $$left|frac{f(x)-f(a)}{g(x)-g(a)}-lright|<varepsilon/2$$ so substituting all these inequalities in the large equation in the answer above gives the term $2varepsilon$. The key is that item (a) holds for any $varepsilon$, so in particular for anything of the form $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:14
1
1
(cont.) How large $x$ needs to be will depend, however, on your given $varepsilon$, $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:17
(cont.) How large $x$ needs to be will depend, however, on your given $varepsilon$, $varepsilon/2$, $varepsilon^2$, etc...
– Luiz Cordeiro
Dec 2 at 0:17
add a comment |
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Well note that by assumption $g'(x) $ is non-zero after a certain value of $x$ say $a$ and by Darboux theorem it is of constant sign. Thus $g$ is strictly monotone in $[a, infty) $ and that means $g(x) neq g(a)$ if $x>a$. The other part is mostly the typical $epsilon, delta$ gymnastics which involves triangle inequality and some simple bounds for various terms.
– Paramanand Singh
Dec 1 at 15:50