Explain why this definition of limit is incorrect
$begingroup$
Let $f:(0,1) to mathbb{R}$ be a given function. Explain how the following definition is not equivalent to the definition of the limit
$limlimits_{x to x_0} f(x) = L$
of $f$ at $x_0 in [0,1]$ .
For any $epsilon gt 0$,for any $delta gt 0$ such that for all $x in (0,1)$ and $0 lt|x-x_0| lt delta$, one has $|f(x) - L| lt epsilon$ .
This definition is incorrect because for any $epsilon gt 0$ there exists some $delta gt 0$ that is small enough. It can't be any delta. Is this the only reason why this definition is not valid?
limits definition epsilon-delta
edited Dec 17 '18 at 13:29
GNUSupporter 8964民主女神 地下教會
13.3k72549
asked Dec 16 '18 at 19:23
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
$endgroup$
add a comment |
$begingroup$
Let $f:(0,1) to mathbb{R}$ be a given function. Explain how the following definition is not equivalent to the definition of the limit
$limlimits_{x to x_0} f(x) = L$
of $f$ at $x_0 in [0,1]$ .
For any $epsilon gt 0$,for any $delta gt 0$ such that for all $x in (0,1)$ and $0 lt|x-x_0| lt delta$, one has $|f(x) - L| lt epsilon$ .
This definition is incorrect because for any $epsilon gt 0$ there exists some $delta gt 0$ that is small enough. It can't be any delta. Is this the only reason why this definition is not valid?
limits definition epsilon-delta
edited Dec 17 '18 at 13:29
GNUSupporter 8964民主女神 地下教會
13.3k72549
asked Dec 16 '18 at 19:23
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
$endgroup$
4
$begingroup$
Please change your username to something less self-insulting??
$endgroup$
– amWhy
Dec 16 '18 at 19:31
add a comment |
$begingroup$
Let $f:(0,1) to mathbb{R}$ be a given function. Explain how the following definition is not equivalent to the definition of the limit
$limlimits_{x to x_0} f(x) = L$
of $f$ at $x_0 in [0,1]$ .
For any $epsilon gt 0$,for any $delta gt 0$ such that for all $x in (0,1)$ and $0 lt|x-x_0| lt delta$, one has $|f(x) - L| lt epsilon$ .
This definition is incorrect because for any $epsilon gt 0$ there exists some $delta gt 0$ that is small enough. It can't be any delta. Is this the only reason why this definition is not valid?
limits definition epsilon-delta
edited Dec 17 '18 at 13:29
GNUSupporter 8964民主女神 地下教會
13.3k72549
asked Dec 16 '18 at 19:23
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
$endgroup$
Let $f:(0,1) to mathbb{R}$ be a given function. Explain how the following definition is not equivalent to the definition of the limit
$limlimits_{x to x_0} f(x) = L$
of $f$ at $x_0 in [0,1]$ .
For any $epsilon gt 0$,for any $delta gt 0$ such that for all $x in (0,1)$ and $0 lt|x-x_0| lt delta$, one has $|f(x) - L| lt epsilon$ .
This definition is incorrect because for any $epsilon gt 0$ there exists some $delta gt 0$ that is small enough. It can't be any delta. Is this the only reason why this definition is not valid?
limits definition epsilon-delta
limits definition epsilon-delta
edited Dec 17 '18 at 13:29
GNUSupporter 8964民主女神 地下教會
13.3k72549
asked Dec 16 '18 at 19:23
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
edited Dec 17 '18 at 13:29
GNUSupporter 8964民主女神 地下教會
13.3k72549
asked Dec 16 '18 at 19:23
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
edited Dec 17 '18 at 13:29
GNUSupporter 8964民主女神 地下教會
13.3k72549
edited Dec 17 '18 at 13:29
GNUSupporter 8964民主女神 地下教會
13.3k72549
edited Dec 17 '18 at 13:29
GNUSupporter 8964民主女神 地下教會
13.3k72549
13.3k72549
asked Dec 16 '18 at 19:23
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
asked Dec 16 '18 at 19:23
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
asked Dec 16 '18 at 19:23
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
1619
4
$begingroup$
Please change your username to something less self-insulting??
$endgroup$
– amWhy
Dec 16 '18 at 19:31
add a comment |
4
$begingroup$
Please change your username to something less self-insulting??
$endgroup$
– amWhy
Dec 16 '18 at 19:31
4
4
$begingroup$
Please change your username to something less self-insulting??
$endgroup$
– amWhy
Dec 16 '18 at 19:31
$begingroup$
Please change your username to something less self-insulting??
$endgroup$
– amWhy
Dec 16 '18 at 19:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes the correct definition requires
$$forall epsilon >0 quad exists delta >0 quad ldots$$
and the other part of the definition is correct.
Indeed let consider for example $f(x)=x$ with $lim_{xto 1} x=1$ and take $epsilon =.01$ and $delta =.5$ then assume $x$ such that $0<|x-1|<0.5$ that is $x=1.4$ and we have
$$|f(x)-1|=|1.4-1|=0.4>epsilon$$
answered Dec 16 '18 at 19:25
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
but could refresh my memory on why it has to be there exists delta greater than 0 instead of any delta greater than 0??
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 19:29
$begingroup$
@ISuckAtMathPleaseHELPME You can consider a simple case for example $f(x)=x$ and try to apply the other definition to see that it doesn't work.
$endgroup$
– gimusi
Dec 16 '18 at 19:32
add a comment |
$begingroup$
Your definition implies the known definition but the converse is not true.
Take $f$ defined by :
$$f(x)=0 ;; text{ if } ;; x<frac 12$$
and
$$f(x)=color{red}{4};; text{ if } ;; xge frac 12.$$
then we have $$lim_{xto 0^+}f(x)=0$$
but
if we take $epsilon = color{red}{3}$
then we do not have
$$forall eta>0 ;; forall xin(0,1)$$
$$|x-0|<eta implies ;; |f(x)-0|<3$$
for example $f(frac{9}{10})=color{red}{4}>epsilon$
answered Dec 16 '18 at 19:32
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
Yes the correct definition requires
$$forall epsilon >0 quad exists delta >0 quad ldots$$
and the other part of the definition is correct.
Indeed let consider for example $f(x)=x$ with $lim_{xto 1} x=1$ and take $epsilon =.01$ and $delta =.5$ then assume $x$ such that $0<|x-1|<0.5$ that is $x=1.4$ and we have
$$|f(x)-1|=|1.4-1|=0.4>epsilon$$
answered Dec 16 '18 at 19:25
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
but could refresh my memory on why it has to be there exists delta greater than 0 instead of any delta greater than 0??
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 19:29
$begingroup$
@ISuckAtMathPleaseHELPME You can consider a simple case for example $f(x)=x$ and try to apply the other definition to see that it doesn't work.
$endgroup$
– gimusi
Dec 16 '18 at 19:32
add a comment |
$begingroup$
Yes the correct definition requires
$$forall epsilon >0 quad exists delta >0 quad ldots$$
and the other part of the definition is correct.
Indeed let consider for example $f(x)=x$ with $lim_{xto 1} x=1$ and take $epsilon =.01$ and $delta =.5$ then assume $x$ such that $0<|x-1|<0.5$ that is $x=1.4$ and we have
$$|f(x)-1|=|1.4-1|=0.4>epsilon$$
answered Dec 16 '18 at 19:25
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
but could refresh my memory on why it has to be there exists delta greater than 0 instead of any delta greater than 0??
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 19:29
$begingroup$
@ISuckAtMathPleaseHELPME You can consider a simple case for example $f(x)=x$ and try to apply the other definition to see that it doesn't work.
$endgroup$
– gimusi
Dec 16 '18 at 19:32
add a comment |
$begingroup$
Yes the correct definition requires
$$forall epsilon >0 quad exists delta >0 quad ldots$$
and the other part of the definition is correct.
Indeed let consider for example $f(x)=x$ with $lim_{xto 1} x=1$ and take $epsilon =.01$ and $delta =.5$ then assume $x$ such that $0<|x-1|<0.5$ that is $x=1.4$ and we have
$$|f(x)-1|=|1.4-1|=0.4>epsilon$$
answered Dec 16 '18 at 19:25
gimusigimusi
92.9k84494
$endgroup$
Yes the correct definition requires
$$forall epsilon >0 quad exists delta >0 quad ldots$$
and the other part of the definition is correct.
Indeed let consider for example $f(x)=x$ with $lim_{xto 1} x=1$ and take $epsilon =.01$ and $delta =.5$ then assume $x$ such that $0<|x-1|<0.5$ that is $x=1.4$ and we have
$$|f(x)-1|=|1.4-1|=0.4>epsilon$$
answered Dec 16 '18 at 19:25
gimusigimusi
92.9k84494
edited Dec 16 '18 at 19:35
answered Dec 16 '18 at 19:25
gimusigimusi
92.9k84494
answered Dec 16 '18 at 19:25
gimusigimusi
92.9k84494
answered Dec 16 '18 at 19:25
gimusigimusi
92.9k84494
92.9k84494
$begingroup$
but could refresh my memory on why it has to be there exists delta greater than 0 instead of any delta greater than 0??
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 19:29
$begingroup$
@ISuckAtMathPleaseHELPME You can consider a simple case for example $f(x)=x$ and try to apply the other definition to see that it doesn't work.
$endgroup$
– gimusi
Dec 16 '18 at 19:32
add a comment |
$begingroup$
but could refresh my memory on why it has to be there exists delta greater than 0 instead of any delta greater than 0??
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 19:29
$begingroup$
@ISuckAtMathPleaseHELPME You can consider a simple case for example $f(x)=x$ and try to apply the other definition to see that it doesn't work.
$endgroup$
– gimusi
Dec 16 '18 at 19:32
$begingroup$
but could refresh my memory on why it has to be there exists delta greater than 0 instead of any delta greater than 0??
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 19:29
$begingroup$
but could refresh my memory on why it has to be there exists delta greater than 0 instead of any delta greater than 0??
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 19:29
$begingroup$
@ISuckAtMathPleaseHELPME You can consider a simple case for example $f(x)=x$ and try to apply the other definition to see that it doesn't work.
$endgroup$
– gimusi
Dec 16 '18 at 19:32
$begingroup$
@ISuckAtMathPleaseHELPME You can consider a simple case for example $f(x)=x$ and try to apply the other definition to see that it doesn't work.
$endgroup$
– gimusi
Dec 16 '18 at 19:32
add a comment |
$begingroup$
Your definition implies the known definition but the converse is not true.
Take $f$ defined by :
$$f(x)=0 ;; text{ if } ;; x<frac 12$$
and
$$f(x)=color{red}{4};; text{ if } ;; xge frac 12.$$
then we have $$lim_{xto 0^+}f(x)=0$$
but
if we take $epsilon = color{red}{3}$
then we do not have
$$forall eta>0 ;; forall xin(0,1)$$
$$|x-0|<eta implies ;; |f(x)-0|<3$$
for example $f(frac{9}{10})=color{red}{4}>epsilon$
answered Dec 16 '18 at 19:32
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
Your definition implies the known definition but the converse is not true.
Take $f$ defined by :
$$f(x)=0 ;; text{ if } ;; x<frac 12$$
and
$$f(x)=color{red}{4};; text{ if } ;; xge frac 12.$$
then we have $$lim_{xto 0^+}f(x)=0$$
but
if we take $epsilon = color{red}{3}$
then we do not have
$$forall eta>0 ;; forall xin(0,1)$$
$$|x-0|<eta implies ;; |f(x)-0|<3$$
for example $f(frac{9}{10})=color{red}{4}>epsilon$
answered Dec 16 '18 at 19:32
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
Your definition implies the known definition but the converse is not true.
Take $f$ defined by :
$$f(x)=0 ;; text{ if } ;; x<frac 12$$
and
$$f(x)=color{red}{4};; text{ if } ;; xge frac 12.$$
then we have $$lim_{xto 0^+}f(x)=0$$
but
if we take $epsilon = color{red}{3}$
then we do not have
$$forall eta>0 ;; forall xin(0,1)$$
$$|x-0|<eta implies ;; |f(x)-0|<3$$
for example $f(frac{9}{10})=color{red}{4}>epsilon$
answered Dec 16 '18 at 19:32
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
Your definition implies the known definition but the converse is not true.
Take $f$ defined by :
$$f(x)=0 ;; text{ if } ;; x<frac 12$$
and
$$f(x)=color{red}{4};; text{ if } ;; xge frac 12.$$
then we have $$lim_{xto 0^+}f(x)=0$$
but
if we take $epsilon = color{red}{3}$
then we do not have
$$forall eta>0 ;; forall xin(0,1)$$
$$|x-0|<eta implies ;; |f(x)-0|<3$$
for example $f(frac{9}{10})=color{red}{4}>epsilon$
answered Dec 16 '18 at 19:32
hamam_Abdallahhamam_Abdallah
38.1k21634
edited Dec 16 '18 at 19:53
answered Dec 16 '18 at 19:32
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 16 '18 at 19:32
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 16 '18 at 19:32
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
add a comment |
add a comment |
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$begingroup$
Please change your username to something less self-insulting??
$endgroup$
– amWhy
Dec 16 '18 at 19:31