Mixing
Problem regarding Epsilon-Delta definition of limits
$begingroup$
I came across this problem in "Introduction to Real Analysis" by Bartle and had a doubt regarding the solution to a given problem.
I needed to determine a condition on $|x-1|$ that will ensure that
$$|(x^2-1)| < 1/2 $$
I started out by factorising : $|x^2-1| = |(x-1)(x+1)| <1/2 $
We have control over $|x-1|$ however we need to get rid of the $|x+1|$ factor to get a relation very similar to $|x-1| <delta$.
This is where I am having a problem. In the examples given, they have generally assumed that $|x-1|<1$. By presupposing this condition, I am able to solve it but am not able to understand why 1 is being chosen since it seems like an arbitrary choice and the method should also work for some other value of $delta$. This does not seem to be the case as if you take any other number, you do not get the correct condition. Could someone please explain the rational behind using only 1? Thank You.
real-analysis calculus limits epsilon-delta
asked Jan 23 at 9:06
Aditya SriramAditya Sriram
132
$endgroup$
add a comment |
$begingroup$
I came across this problem in "Introduction to Real Analysis" by Bartle and had a doubt regarding the solution to a given problem.
I needed to determine a condition on $|x-1|$ that will ensure that
$$|(x^2-1)| < 1/2 $$
I started out by factorising : $|x^2-1| = |(x-1)(x+1)| <1/2 $
We have control over $|x-1|$ however we need to get rid of the $|x+1|$ factor to get a relation very similar to $|x-1| <delta$.
This is where I am having a problem. In the examples given, they have generally assumed that $|x-1|<1$. By presupposing this condition, I am able to solve it but am not able to understand why 1 is being chosen since it seems like an arbitrary choice and the method should also work for some other value of $delta$. This does not seem to be the case as if you take any other number, you do not get the correct condition. Could someone please explain the rational behind using only 1? Thank You.
real-analysis calculus limits epsilon-delta
asked Jan 23 at 9:06
Aditya SriramAditya Sriram
132
$endgroup$
$begingroup$
Such $epsilon, delta$ questions do not have a unique answer. Your $delta$ may be different from your peer's and both answers may be correct.
$endgroup$
– Paramanand Singh
Jan 24 at 3:37
$begingroup$
@ParamanandSingh Alright thank you because this is usually what I have struggled in Epsilon-Delta proofs.
$endgroup$
– Aditya Sriram
Jan 24 at 4:10
add a comment |
$begingroup$
I came across this problem in "Introduction to Real Analysis" by Bartle and had a doubt regarding the solution to a given problem.
I needed to determine a condition on $|x-1|$ that will ensure that
$$|(x^2-1)| < 1/2 $$
I started out by factorising : $|x^2-1| = |(x-1)(x+1)| <1/2 $
We have control over $|x-1|$ however we need to get rid of the $|x+1|$ factor to get a relation very similar to $|x-1| <delta$.
This is where I am having a problem. In the examples given, they have generally assumed that $|x-1|<1$. By presupposing this condition, I am able to solve it but am not able to understand why 1 is being chosen since it seems like an arbitrary choice and the method should also work for some other value of $delta$. This does not seem to be the case as if you take any other number, you do not get the correct condition. Could someone please explain the rational behind using only 1? Thank You.
real-analysis calculus limits epsilon-delta
asked Jan 23 at 9:06
Aditya SriramAditya Sriram
132
$endgroup$
I came across this problem in "Introduction to Real Analysis" by Bartle and had a doubt regarding the solution to a given problem.
I needed to determine a condition on $|x-1|$ that will ensure that
$$|(x^2-1)| < 1/2 $$
I started out by factorising : $|x^2-1| = |(x-1)(x+1)| <1/2 $
We have control over $|x-1|$ however we need to get rid of the $|x+1|$ factor to get a relation very similar to $|x-1| <delta$.
This is where I am having a problem. In the examples given, they have generally assumed that $|x-1|<1$. By presupposing this condition, I am able to solve it but am not able to understand why 1 is being chosen since it seems like an arbitrary choice and the method should also work for some other value of $delta$. This does not seem to be the case as if you take any other number, you do not get the correct condition. Could someone please explain the rational behind using only 1? Thank You.
real-analysis calculus limits epsilon-delta
real-analysis calculus limits epsilon-delta
asked Jan 23 at 9:06
Aditya SriramAditya Sriram
132
asked Jan 23 at 9:06
Aditya SriramAditya Sriram
132
asked Jan 23 at 9:06
Aditya SriramAditya Sriram
132
asked Jan 23 at 9:06
Aditya SriramAditya Sriram
132
asked Jan 23 at 9:06
Aditya SriramAditya Sriram
132
132
$begingroup$
Such $epsilon, delta$ questions do not have a unique answer. Your $delta$ may be different from your peer's and both answers may be correct.
$endgroup$
– Paramanand Singh
Jan 24 at 3:37
$begingroup$
@ParamanandSingh Alright thank you because this is usually what I have struggled in Epsilon-Delta proofs.
$endgroup$
– Aditya Sriram
Jan 24 at 4:10
add a comment |
$begingroup$
Such $epsilon, delta$ questions do not have a unique answer. Your $delta$ may be different from your peer's and both answers may be correct.
$endgroup$
– Paramanand Singh
Jan 24 at 3:37
$begingroup$
@ParamanandSingh Alright thank you because this is usually what I have struggled in Epsilon-Delta proofs.
$endgroup$
– Aditya Sriram
Jan 24 at 4:10
$begingroup$
Such $epsilon, delta$ questions do not have a unique answer. Your $delta$ may be different from your peer's and both answers may be correct.
$endgroup$
– Paramanand Singh
Jan 24 at 3:37
$begingroup$
Such $epsilon, delta$ questions do not have a unique answer. Your $delta$ may be different from your peer's and both answers may be correct.
$endgroup$
– Paramanand Singh
Jan 24 at 3:37
$begingroup$
@ParamanandSingh Alright thank you because this is usually what I have struggled in Epsilon-Delta proofs.
$endgroup$
– Aditya Sriram
Jan 24 at 4:10
$begingroup$
@ParamanandSingh Alright thank you because this is usually what I have struggled in Epsilon-Delta proofs.
$endgroup$
– Aditya Sriram
Jan 24 at 4:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
People usually choose $1$ because that's the simplest choice. But any number greater than $0$ will do. I will do it with $4$. So, let us start by assuming that $lvert x-1rvert<4$. Thenbegin{align}lvert x+1rvert&=bigllvert(x-1)+2bigrrvert\&leqslantlvert x-1rvert+2\&<6.end{align}So, for any $varepsilon>0$, take $delta=minleft{4,fracvarepsilon6right}$, and thenbegin{align}lvert x-1rvert<deltaimplieslvert x^2-1rvert&=lvert x-1rverttimeslvert x+1rvert\&<fracvarepsilon6times6\&=varepsilon.end{align}
answered Jan 23 at 9:17
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
Does this imply that the approach is equally valid for any value of $delta$? Also, if we consider $delta$ to be less than 1, we get a wider range for the values of |x-1| while if $delta$ is greater than 1 then the range is limited.
$endgroup$
– Aditya Sriram
Jan 23 at 9:31
$begingroup$
I don't understand your question. The number $delta$ is not any number. In my approach, it belongs to $left(0,minleft{4,fracvarepsilon6right}right)$.
$endgroup$
– José Carlos Santos
Jan 23 at 9:33
$begingroup$
Sorry I meant to ask if it is alright to consider
$endgroup$
– Aditya Sriram
Jan 23 at 9:37
$begingroup$
And what does that mean?
$endgroup$
– José Carlos Santos
Jan 23 at 9:43
$begingroup$
Sorry I meant to ask if it is alright to consider 0<|x-1|<1 because for such values, we get an extended range. For eg: if |x-1|<1/2, then we get |x+1|<5/2 which will imply that |x-1|<$epsilon$/5. Therefore we get $delta$ belongs to (0, min{1/2,$epsilon$/5}).
$endgroup$
– Aditya Sriram
Jan 23 at 9:51
|
show 2 more comments
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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oldest
votes
$begingroup$
People usually choose $1$ because that's the simplest choice. But any number greater than $0$ will do. I will do it with $4$. So, let us start by assuming that $lvert x-1rvert<4$. Thenbegin{align}lvert x+1rvert&=bigllvert(x-1)+2bigrrvert\&leqslantlvert x-1rvert+2\&<6.end{align}So, for any $varepsilon>0$, take $delta=minleft{4,fracvarepsilon6right}$, and thenbegin{align}lvert x-1rvert<deltaimplieslvert x^2-1rvert&=lvert x-1rverttimeslvert x+1rvert\&<fracvarepsilon6times6\&=varepsilon.end{align}
answered Jan 23 at 9:17
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
Does this imply that the approach is equally valid for any value of $delta$? Also, if we consider $delta$ to be less than 1, we get a wider range for the values of |x-1| while if $delta$ is greater than 1 then the range is limited.
$endgroup$
– Aditya Sriram
Jan 23 at 9:31
$begingroup$
I don't understand your question. The number $delta$ is not any number. In my approach, it belongs to $left(0,minleft{4,fracvarepsilon6right}right)$.
$endgroup$
– José Carlos Santos
Jan 23 at 9:33
$begingroup$
Sorry I meant to ask if it is alright to consider
$endgroup$
– Aditya Sriram
Jan 23 at 9:37
$begingroup$
And what does that mean?
$endgroup$
– José Carlos Santos
Jan 23 at 9:43
$begingroup$
Sorry I meant to ask if it is alright to consider 0<|x-1|<1 because for such values, we get an extended range. For eg: if |x-1|<1/2, then we get |x+1|<5/2 which will imply that |x-1|<$epsilon$/5. Therefore we get $delta$ belongs to (0, min{1/2,$epsilon$/5}).
$endgroup$
– Aditya Sriram
Jan 23 at 9:51
|
show 2 more comments
$begingroup$
People usually choose $1$ because that's the simplest choice. But any number greater than $0$ will do. I will do it with $4$. So, let us start by assuming that $lvert x-1rvert<4$. Thenbegin{align}lvert x+1rvert&=bigllvert(x-1)+2bigrrvert\&leqslantlvert x-1rvert+2\&<6.end{align}So, for any $varepsilon>0$, take $delta=minleft{4,fracvarepsilon6right}$, and thenbegin{align}lvert x-1rvert<deltaimplieslvert x^2-1rvert&=lvert x-1rverttimeslvert x+1rvert\&<fracvarepsilon6times6\&=varepsilon.end{align}
answered Jan 23 at 9:17
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
Does this imply that the approach is equally valid for any value of $delta$? Also, if we consider $delta$ to be less than 1, we get a wider range for the values of |x-1| while if $delta$ is greater than 1 then the range is limited.
$endgroup$
– Aditya Sriram
Jan 23 at 9:31
$begingroup$
I don't understand your question. The number $delta$ is not any number. In my approach, it belongs to $left(0,minleft{4,fracvarepsilon6right}right)$.
$endgroup$
– José Carlos Santos
Jan 23 at 9:33
$begingroup$
Sorry I meant to ask if it is alright to consider
$endgroup$
– Aditya Sriram
Jan 23 at 9:37
$begingroup$
And what does that mean?
$endgroup$
– José Carlos Santos
Jan 23 at 9:43
$begingroup$
Sorry I meant to ask if it is alright to consider 0<|x-1|<1 because for such values, we get an extended range. For eg: if |x-1|<1/2, then we get |x+1|<5/2 which will imply that |x-1|<$epsilon$/5. Therefore we get $delta$ belongs to (0, min{1/2,$epsilon$/5}).
$endgroup$
– Aditya Sriram
Jan 23 at 9:51
|
show 2 more comments
$begingroup$
People usually choose $1$ because that's the simplest choice. But any number greater than $0$ will do. I will do it with $4$. So, let us start by assuming that $lvert x-1rvert<4$. Thenbegin{align}lvert x+1rvert&=bigllvert(x-1)+2bigrrvert\&leqslantlvert x-1rvert+2\&<6.end{align}So, for any $varepsilon>0$, take $delta=minleft{4,fracvarepsilon6right}$, and thenbegin{align}lvert x-1rvert<deltaimplieslvert x^2-1rvert&=lvert x-1rverttimeslvert x+1rvert\&<fracvarepsilon6times6\&=varepsilon.end{align}
answered Jan 23 at 9:17
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
People usually choose $1$ because that's the simplest choice. But any number greater than $0$ will do. I will do it with $4$. So, let us start by assuming that $lvert x-1rvert<4$. Thenbegin{align}lvert x+1rvert&=bigllvert(x-1)+2bigrrvert\&leqslantlvert x-1rvert+2\&<6.end{align}So, for any $varepsilon>0$, take $delta=minleft{4,fracvarepsilon6right}$, and thenbegin{align}lvert x-1rvert<deltaimplieslvert x^2-1rvert&=lvert x-1rverttimeslvert x+1rvert\&<fracvarepsilon6times6\&=varepsilon.end{align}
answered Jan 23 at 9:17
José Carlos SantosJosé Carlos Santos
167k22132235
edited Jan 23 at 9:23
answered Jan 23 at 9:17
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 23 at 9:17
José Carlos SantosJosé Carlos Santos
167k22132235
answered Jan 23 at 9:17
José Carlos SantosJosé Carlos Santos
167k22132235
167k22132235
$begingroup$
Does this imply that the approach is equally valid for any value of $delta$? Also, if we consider $delta$ to be less than 1, we get a wider range for the values of |x-1| while if $delta$ is greater than 1 then the range is limited.
$endgroup$
– Aditya Sriram
Jan 23 at 9:31
$begingroup$
I don't understand your question. The number $delta$ is not any number. In my approach, it belongs to $left(0,minleft{4,fracvarepsilon6right}right)$.
$endgroup$
– José Carlos Santos
Jan 23 at 9:33
$begingroup$
Sorry I meant to ask if it is alright to consider
$endgroup$
– Aditya Sriram
Jan 23 at 9:37
$begingroup$
And what does that mean?
$endgroup$
– José Carlos Santos
Jan 23 at 9:43
$begingroup$
Sorry I meant to ask if it is alright to consider 0<|x-1|<1 because for such values, we get an extended range. For eg: if |x-1|<1/2, then we get |x+1|<5/2 which will imply that |x-1|<$epsilon$/5. Therefore we get $delta$ belongs to (0, min{1/2,$epsilon$/5}).
$endgroup$
– Aditya Sriram
Jan 23 at 9:51
|
show 2 more comments
$begingroup$
Does this imply that the approach is equally valid for any value of $delta$? Also, if we consider $delta$ to be less than 1, we get a wider range for the values of |x-1| while if $delta$ is greater than 1 then the range is limited.
$endgroup$
– Aditya Sriram
Jan 23 at 9:31
$begingroup$
I don't understand your question. The number $delta$ is not any number. In my approach, it belongs to $left(0,minleft{4,fracvarepsilon6right}right)$.
$endgroup$
– José Carlos Santos
Jan 23 at 9:33
$begingroup$
Sorry I meant to ask if it is alright to consider
$endgroup$
– Aditya Sriram
Jan 23 at 9:37
$begingroup$
And what does that mean?
$endgroup$
– José Carlos Santos
Jan 23 at 9:43
$begingroup$
Sorry I meant to ask if it is alright to consider 0<|x-1|<1 because for such values, we get an extended range. For eg: if |x-1|<1/2, then we get |x+1|<5/2 which will imply that |x-1|<$epsilon$/5. Therefore we get $delta$ belongs to (0, min{1/2,$epsilon$/5}).
$endgroup$
– Aditya Sriram
Jan 23 at 9:51
$begingroup$
Does this imply that the approach is equally valid for any value of $delta$? Also, if we consider $delta$ to be less than 1, we get a wider range for the values of |x-1| while if $delta$ is greater than 1 then the range is limited.
$endgroup$
– Aditya Sriram
Jan 23 at 9:31
$begingroup$
Does this imply that the approach is equally valid for any value of $delta$? Also, if we consider $delta$ to be less than 1, we get a wider range for the values of |x-1| while if $delta$ is greater than 1 then the range is limited.
$endgroup$
– Aditya Sriram
Jan 23 at 9:31
$begingroup$
I don't understand your question. The number $delta$ is not any number. In my approach, it belongs to $left(0,minleft{4,fracvarepsilon6right}right)$.
$endgroup$
– José Carlos Santos
Jan 23 at 9:33
$begingroup$
I don't understand your question. The number $delta$ is not any number. In my approach, it belongs to $left(0,minleft{4,fracvarepsilon6right}right)$.
$endgroup$
– José Carlos Santos
Jan 23 at 9:33
$begingroup$
Sorry I meant to ask if it is alright to consider
$endgroup$
– Aditya Sriram
Jan 23 at 9:37
$begingroup$
Sorry I meant to ask if it is alright to consider
$endgroup$
– Aditya Sriram
Jan 23 at 9:37
$begingroup$
And what does that mean?
$endgroup$
– José Carlos Santos
Jan 23 at 9:43
$begingroup$
And what does that mean?
$endgroup$
– José Carlos Santos
Jan 23 at 9:43
$begingroup$
Sorry I meant to ask if it is alright to consider 0<|x-1|<1 because for such values, we get an extended range. For eg: if |x-1|<1/2, then we get |x+1|<5/2 which will imply that |x-1|<$epsilon$/5. Therefore we get $delta$ belongs to (0, min{1/2,$epsilon$/5}).
$endgroup$
– Aditya Sriram
Jan 23 at 9:51
$begingroup$
Sorry I meant to ask if it is alright to consider 0<|x-1|<1 because for such values, we get an extended range. For eg: if |x-1|<1/2, then we get |x+1|<5/2 which will imply that |x-1|<$epsilon$/5. Therefore we get $delta$ belongs to (0, min{1/2,$epsilon$/5}).
$endgroup$
– Aditya Sriram
Jan 23 at 9:51
|
show 2 more comments
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$begingroup$
Such $epsilon, delta$ questions do not have a unique answer. Your $delta$ may be different from your peer's and both answers may be correct.
$endgroup$
– Paramanand Singh
Jan 24 at 3:37
$begingroup$
@ParamanandSingh Alright thank you because this is usually what I have struggled in Epsilon-Delta proofs.
$endgroup$
– Aditya Sriram
Jan 24 at 4:10