Mixing
How can I continue this proof?
$begingroup$
I am trying to prove the theorem of limit of composition of functions under the scenario:
$lim_{xto a}f(x)=l$ , $lim_{xto b}g(x)=a$. I need to prove $lim_{xto b}f(g(x))=l$.
There are two cases where this is valid:
(i) If $f$ is continuous at $a$
(ii) $f$ is not continuous at $a$, but there exists an open interval $I$ containing $b$ such that $g(x)neq a$ $forall xin I $ except possibly at $b$.
I made a proof for (i) easily. But have a problem in (ii)
My proof(incomplete) for (ii) :
Let $epsilon>0$
Hence, $existsdelta>0$
such that |u-a|<$delta$, u$neq a$ then |f(u)-l|<$epsilon$ .
Now $g(x)neq a$ $forall xin I $ except possibly at $ b$. There exists h>0 such that |x-b|<$h$, x$neq$b then
|g(x)-a|<$delta$ , as $delta$>0. But we have no guarantee that |g(x)-a|$neq$0 (We require that to put u=g(x))
Here's where I am stuck. I can't just say that $h$ is such a number that |x-b|<$h$, x$neq$b then
0<|g(x)-a|<$delta$, because that is an extra restriction on |g(x)-a|. The limit definition would only give |g(x)-a|<$delta$ for sufficiently small h>0, but |g(x)-a|$neq$0 is not necessary. Simply, I have to prove that there always exists $h>0$ , such that $(b-h,b+h)$ is a subset of $I$ and in that interval, $0<|g(x)-a|< delta $ (see that "there always exists" part is what I need to prove)
Is this assertion true? If so, why?
calculus limits
asked Sep 4 '18 at 17:21
SteveSteve
828
$endgroup$
add a comment |
$begingroup$
I am trying to prove the theorem of limit of composition of functions under the scenario:
$lim_{xto a}f(x)=l$ , $lim_{xto b}g(x)=a$. I need to prove $lim_{xto b}f(g(x))=l$.
There are two cases where this is valid:
(i) If $f$ is continuous at $a$
(ii) $f$ is not continuous at $a$, but there exists an open interval $I$ containing $b$ such that $g(x)neq a$ $forall xin I $ except possibly at $b$.
I made a proof for (i) easily. But have a problem in (ii)
My proof(incomplete) for (ii) :
Let $epsilon>0$
Hence, $existsdelta>0$
such that |u-a|<$delta$, u$neq a$ then |f(u)-l|<$epsilon$ .
Now $g(x)neq a$ $forall xin I $ except possibly at $ b$. There exists h>0 such that |x-b|<$h$, x$neq$b then
|g(x)-a|<$delta$ , as $delta$>0. But we have no guarantee that |g(x)-a|$neq$0 (We require that to put u=g(x))
Here's where I am stuck. I can't just say that $h$ is such a number that |x-b|<$h$, x$neq$b then
0<|g(x)-a|<$delta$, because that is an extra restriction on |g(x)-a|. The limit definition would only give |g(x)-a|<$delta$ for sufficiently small h>0, but |g(x)-a|$neq$0 is not necessary. Simply, I have to prove that there always exists $h>0$ , such that $(b-h,b+h)$ is a subset of $I$ and in that interval, $0<|g(x)-a|< delta $ (see that "there always exists" part is what I need to prove)
Is this assertion true? If so, why?
calculus limits
asked Sep 4 '18 at 17:21
SteveSteve
828
$endgroup$
$begingroup$
Your claim (ii) isn't correct in the first place: for all you know, $f$ can be discontinuous at $a$ and $g$ can be constantly equal to $a$, or be $g(x)=a+(x-b)sinfrac1{x-b}$.
$endgroup$
– Saucy O'Path
Sep 4 '18 at 17:27
$begingroup$
But that's what is exactly covered in (ii) "If f is discontinuous at a, then there exists an interval containing b such that g(x) $neq$ a for all x in I except possibly at b" which means that g(x) is not constantly equal to a in an interval around b.
$endgroup$
– Steve
Sep 4 '18 at 17:34
add a comment |
$begingroup$
I am trying to prove the theorem of limit of composition of functions under the scenario:
$lim_{xto a}f(x)=l$ , $lim_{xto b}g(x)=a$. I need to prove $lim_{xto b}f(g(x))=l$.
There are two cases where this is valid:
(i) If $f$ is continuous at $a$
(ii) $f$ is not continuous at $a$, but there exists an open interval $I$ containing $b$ such that $g(x)neq a$ $forall xin I $ except possibly at $b$.
I made a proof for (i) easily. But have a problem in (ii)
My proof(incomplete) for (ii) :
Let $epsilon>0$
Hence, $existsdelta>0$
such that |u-a|<$delta$, u$neq a$ then |f(u)-l|<$epsilon$ .
Now $g(x)neq a$ $forall xin I $ except possibly at $ b$. There exists h>0 such that |x-b|<$h$, x$neq$b then
|g(x)-a|<$delta$ , as $delta$>0. But we have no guarantee that |g(x)-a|$neq$0 (We require that to put u=g(x))
Here's where I am stuck. I can't just say that $h$ is such a number that |x-b|<$h$, x$neq$b then
0<|g(x)-a|<$delta$, because that is an extra restriction on |g(x)-a|. The limit definition would only give |g(x)-a|<$delta$ for sufficiently small h>0, but |g(x)-a|$neq$0 is not necessary. Simply, I have to prove that there always exists $h>0$ , such that $(b-h,b+h)$ is a subset of $I$ and in that interval, $0<|g(x)-a|< delta $ (see that "there always exists" part is what I need to prove)
Is this assertion true? If so, why?
calculus limits
asked Sep 4 '18 at 17:21
SteveSteve
828
$endgroup$
I am trying to prove the theorem of limit of composition of functions under the scenario:
$lim_{xto a}f(x)=l$ , $lim_{xto b}g(x)=a$. I need to prove $lim_{xto b}f(g(x))=l$.
There are two cases where this is valid:
(i) If $f$ is continuous at $a$
(ii) $f$ is not continuous at $a$, but there exists an open interval $I$ containing $b$ such that $g(x)neq a$ $forall xin I $ except possibly at $b$.
I made a proof for (i) easily. But have a problem in (ii)
My proof(incomplete) for (ii) :
Let $epsilon>0$
Hence, $existsdelta>0$
such that |u-a|<$delta$, u$neq a$ then |f(u)-l|<$epsilon$ .
Now $g(x)neq a$ $forall xin I $ except possibly at $ b$. There exists h>0 such that |x-b|<$h$, x$neq$b then
|g(x)-a|<$delta$ , as $delta$>0. But we have no guarantee that |g(x)-a|$neq$0 (We require that to put u=g(x))
Here's where I am stuck. I can't just say that $h$ is such a number that |x-b|<$h$, x$neq$b then
0<|g(x)-a|<$delta$, because that is an extra restriction on |g(x)-a|. The limit definition would only give |g(x)-a|<$delta$ for sufficiently small h>0, but |g(x)-a|$neq$0 is not necessary. Simply, I have to prove that there always exists $h>0$ , such that $(b-h,b+h)$ is a subset of $I$ and in that interval, $0<|g(x)-a|< delta $ (see that "there always exists" part is what I need to prove)
Is this assertion true? If so, why?
calculus limits
calculus limits
asked Sep 4 '18 at 17:21
SteveSteve
828
asked Sep 4 '18 at 17:21
SteveSteve
828
edited Jan 6 at 9:35
Steve
asked Sep 4 '18 at 17:21
SteveSteve
828
asked Sep 4 '18 at 17:21
SteveSteve
828
asked Sep 4 '18 at 17:21
SteveSteve
828
828
$begingroup$
Your claim (ii) isn't correct in the first place: for all you know, $f$ can be discontinuous at $a$ and $g$ can be constantly equal to $a$, or be $g(x)=a+(x-b)sinfrac1{x-b}$.
$endgroup$
– Saucy O'Path
Sep 4 '18 at 17:27
$begingroup$
But that's what is exactly covered in (ii) "If f is discontinuous at a, then there exists an interval containing b such that g(x) $neq$ a for all x in I except possibly at b" which means that g(x) is not constantly equal to a in an interval around b.
$endgroup$
– Steve
Sep 4 '18 at 17:34
add a comment |
$begingroup$
Your claim (ii) isn't correct in the first place: for all you know, $f$ can be discontinuous at $a$ and $g$ can be constantly equal to $a$, or be $g(x)=a+(x-b)sinfrac1{x-b}$.
$endgroup$
– Saucy O'Path
Sep 4 '18 at 17:27
$begingroup$
But that's what is exactly covered in (ii) "If f is discontinuous at a, then there exists an interval containing b such that g(x) $neq$ a for all x in I except possibly at b" which means that g(x) is not constantly equal to a in an interval around b.
$endgroup$
– Steve
Sep 4 '18 at 17:34
$begingroup$
Your claim (ii) isn't correct in the first place: for all you know, $f$ can be discontinuous at $a$ and $g$ can be constantly equal to $a$, or be $g(x)=a+(x-b)sinfrac1{x-b}$.
$endgroup$
– Saucy O'Path
Sep 4 '18 at 17:27
$begingroup$
Your claim (ii) isn't correct in the first place: for all you know, $f$ can be discontinuous at $a$ and $g$ can be constantly equal to $a$, or be $g(x)=a+(x-b)sinfrac1{x-b}$.
$endgroup$
– Saucy O'Path
Sep 4 '18 at 17:27
$begingroup$
But that's what is exactly covered in (ii) "If f is discontinuous at a, then there exists an interval containing b such that g(x) $neq$ a for all x in I except possibly at b" which means that g(x) is not constantly equal to a in an interval around b.
$endgroup$
– Steve
Sep 4 '18 at 17:34
$begingroup$
But that's what is exactly covered in (ii) "If f is discontinuous at a, then there exists an interval containing b such that g(x) $neq$ a for all x in I except possibly at b" which means that g(x) is not constantly equal to a in an interval around b.
$endgroup$
– Steve
Sep 4 '18 at 17:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your statement (ii) is correct: If $lim_{yto a} f(y)=ell$, $ lim_{xto b} g(x)=a$, and $g$ does not assume the value $a$ in a punctured neighborhood $dot U$ of $b$ then $lim_{xto b}fbigl(g(x)bigr)=ell$.
Proof. Let an $epsilon>0$ be given. Then there is a punctured neighborhood $dot V$ of $a$ such that $|f(y)-ell|<epsilon$ for all $yindot V$. Furthermore there is a punctured neighborhood $dot W$ of $b$ such that $g(x)in V$ for all $bindot W$. The set $dot Omega:=dot Wcapdot U$ is a punctured neighborhood of $b$ as well, and for all $xin dot Omega$ we have $g(x)ne a$, hence $g(x)in dot V$. This then implies that
$$bigl|fbigl(g(x)bigr)-ellbigr|<epsilonqquadforall xindotOmega ,$$
and proves the claim. Note that this proof also works when $a$ or $b$ are $ =pminfty$.
In $epsilon/delta$ terms exactly the same proof is much more cumbersome:
Let an $epsilon>0$ be given. Then there is a $delta>0$ such that
$$0<|y-a|<deltaquadRightarrowquad |f(y)-ell|<epsilon .tag{1}$$
Furthermore there is an $eta'>0$ such that
$$0<|x-b|<eta'quadRightarrowquad |g(x)-a|<delta .tag{2}$$
Since $g$ does not assume the value $a$ in a punctured neighborhood of $b$ there is an $eta''>0$ such that $$0<|x-b|<eta''quadRightarrowquad g(x)ne a .tag{3}$$
Put $eta:=min{eta', eta''}>0$. Then $(2)$ and $(3)$ together imply
$$0<|x-b|<etaquadRightarrowquad 0<|g(x)-a|<delta .tag{4}$$
Taking $(4)$ and letting $y:=g(x)$ in $(1)$ we finally obtain
$$0<|x-b|<etaquadRightarrowquad bigl|fbigl(g(x)bigr)-ellbigr|<epsilon .$$
answered Sep 4 '18 at 19:16
Christian BlatterChristian Blatter
173k7113326
$endgroup$
$begingroup$
Can you write your last statement (|f(g(x)-l| < $epsilon$ $forall$ x$in$ $Omega$) in the form of $epsilon$ $delta$ statement, because I am still confused if it satisfies the limit definition ($epsilon$ $delta$) or if that is too long just give the reason why this will be true for all $epsilon$>0 . Thanks for the answer
$endgroup$
– Steve
Sep 5 '18 at 11:05
$begingroup$
can we say that existence of the interval $Omega$ is always guaranteed because the existence of W is always guaranteed ? So,there always exists an interval of b such that |f(g(x))-l|<$epsilon$ , right?
$endgroup$
– Steve
Sep 5 '18 at 11:19
$begingroup$
Thanks for the edit. Now it is crystal clear.
$endgroup$
– Steve
Sep 6 '18 at 15:11
add a comment |
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$begingroup$
Your statement (ii) is correct: If $lim_{yto a} f(y)=ell$, $ lim_{xto b} g(x)=a$, and $g$ does not assume the value $a$ in a punctured neighborhood $dot U$ of $b$ then $lim_{xto b}fbigl(g(x)bigr)=ell$.
Proof. Let an $epsilon>0$ be given. Then there is a punctured neighborhood $dot V$ of $a$ such that $|f(y)-ell|<epsilon$ for all $yindot V$. Furthermore there is a punctured neighborhood $dot W$ of $b$ such that $g(x)in V$ for all $bindot W$. The set $dot Omega:=dot Wcapdot U$ is a punctured neighborhood of $b$ as well, and for all $xin dot Omega$ we have $g(x)ne a$, hence $g(x)in dot V$. This then implies that
$$bigl|fbigl(g(x)bigr)-ellbigr|<epsilonqquadforall xindotOmega ,$$
and proves the claim. Note that this proof also works when $a$ or $b$ are $ =pminfty$.
In $epsilon/delta$ terms exactly the same proof is much more cumbersome:
Let an $epsilon>0$ be given. Then there is a $delta>0$ such that
$$0<|y-a|<deltaquadRightarrowquad |f(y)-ell|<epsilon .tag{1}$$
Furthermore there is an $eta'>0$ such that
$$0<|x-b|<eta'quadRightarrowquad |g(x)-a|<delta .tag{2}$$
Since $g$ does not assume the value $a$ in a punctured neighborhood of $b$ there is an $eta''>0$ such that $$0<|x-b|<eta''quadRightarrowquad g(x)ne a .tag{3}$$
Put $eta:=min{eta', eta''}>0$. Then $(2)$ and $(3)$ together imply
$$0<|x-b|<etaquadRightarrowquad 0<|g(x)-a|<delta .tag{4}$$
Taking $(4)$ and letting $y:=g(x)$ in $(1)$ we finally obtain
$$0<|x-b|<etaquadRightarrowquad bigl|fbigl(g(x)bigr)-ellbigr|<epsilon .$$
answered Sep 4 '18 at 19:16
Christian BlatterChristian Blatter
173k7113326
$endgroup$
$begingroup$
Can you write your last statement (|f(g(x)-l| < $epsilon$ $forall$ x$in$ $Omega$) in the form of $epsilon$ $delta$ statement, because I am still confused if it satisfies the limit definition ($epsilon$ $delta$) or if that is too long just give the reason why this will be true for all $epsilon$>0 . Thanks for the answer
$endgroup$
– Steve
Sep 5 '18 at 11:05
$begingroup$
can we say that existence of the interval $Omega$ is always guaranteed because the existence of W is always guaranteed ? So,there always exists an interval of b such that |f(g(x))-l|<$epsilon$ , right?
$endgroup$
– Steve
Sep 5 '18 at 11:19
$begingroup$
Thanks for the edit. Now it is crystal clear.
$endgroup$
– Steve
Sep 6 '18 at 15:11
add a comment |
$begingroup$
Your statement (ii) is correct: If $lim_{yto a} f(y)=ell$, $ lim_{xto b} g(x)=a$, and $g$ does not assume the value $a$ in a punctured neighborhood $dot U$ of $b$ then $lim_{xto b}fbigl(g(x)bigr)=ell$.
Proof. Let an $epsilon>0$ be given. Then there is a punctured neighborhood $dot V$ of $a$ such that $|f(y)-ell|<epsilon$ for all $yindot V$. Furthermore there is a punctured neighborhood $dot W$ of $b$ such that $g(x)in V$ for all $bindot W$. The set $dot Omega:=dot Wcapdot U$ is a punctured neighborhood of $b$ as well, and for all $xin dot Omega$ we have $g(x)ne a$, hence $g(x)in dot V$. This then implies that
$$bigl|fbigl(g(x)bigr)-ellbigr|<epsilonqquadforall xindotOmega ,$$
and proves the claim. Note that this proof also works when $a$ or $b$ are $ =pminfty$.
In $epsilon/delta$ terms exactly the same proof is much more cumbersome:
Let an $epsilon>0$ be given. Then there is a $delta>0$ such that
$$0<|y-a|<deltaquadRightarrowquad |f(y)-ell|<epsilon .tag{1}$$
Furthermore there is an $eta'>0$ such that
$$0<|x-b|<eta'quadRightarrowquad |g(x)-a|<delta .tag{2}$$
Since $g$ does not assume the value $a$ in a punctured neighborhood of $b$ there is an $eta''>0$ such that $$0<|x-b|<eta''quadRightarrowquad g(x)ne a .tag{3}$$
Put $eta:=min{eta', eta''}>0$. Then $(2)$ and $(3)$ together imply
$$0<|x-b|<etaquadRightarrowquad 0<|g(x)-a|<delta .tag{4}$$
Taking $(4)$ and letting $y:=g(x)$ in $(1)$ we finally obtain
$$0<|x-b|<etaquadRightarrowquad bigl|fbigl(g(x)bigr)-ellbigr|<epsilon .$$
answered Sep 4 '18 at 19:16
Christian BlatterChristian Blatter
173k7113326
$endgroup$
$begingroup$
Can you write your last statement (|f(g(x)-l| < $epsilon$ $forall$ x$in$ $Omega$) in the form of $epsilon$ $delta$ statement, because I am still confused if it satisfies the limit definition ($epsilon$ $delta$) or if that is too long just give the reason why this will be true for all $epsilon$>0 . Thanks for the answer
$endgroup$
– Steve
Sep 5 '18 at 11:05
$begingroup$
can we say that existence of the interval $Omega$ is always guaranteed because the existence of W is always guaranteed ? So,there always exists an interval of b such that |f(g(x))-l|<$epsilon$ , right?
$endgroup$
– Steve
Sep 5 '18 at 11:19
$begingroup$
Thanks for the edit. Now it is crystal clear.
$endgroup$
– Steve
Sep 6 '18 at 15:11
add a comment |
$begingroup$
Your statement (ii) is correct: If $lim_{yto a} f(y)=ell$, $ lim_{xto b} g(x)=a$, and $g$ does not assume the value $a$ in a punctured neighborhood $dot U$ of $b$ then $lim_{xto b}fbigl(g(x)bigr)=ell$.
Proof. Let an $epsilon>0$ be given. Then there is a punctured neighborhood $dot V$ of $a$ such that $|f(y)-ell|<epsilon$ for all $yindot V$. Furthermore there is a punctured neighborhood $dot W$ of $b$ such that $g(x)in V$ for all $bindot W$. The set $dot Omega:=dot Wcapdot U$ is a punctured neighborhood of $b$ as well, and for all $xin dot Omega$ we have $g(x)ne a$, hence $g(x)in dot V$. This then implies that
$$bigl|fbigl(g(x)bigr)-ellbigr|<epsilonqquadforall xindotOmega ,$$
and proves the claim. Note that this proof also works when $a$ or $b$ are $ =pminfty$.
In $epsilon/delta$ terms exactly the same proof is much more cumbersome:
Let an $epsilon>0$ be given. Then there is a $delta>0$ such that
$$0<|y-a|<deltaquadRightarrowquad |f(y)-ell|<epsilon .tag{1}$$
Furthermore there is an $eta'>0$ such that
$$0<|x-b|<eta'quadRightarrowquad |g(x)-a|<delta .tag{2}$$
Since $g$ does not assume the value $a$ in a punctured neighborhood of $b$ there is an $eta''>0$ such that $$0<|x-b|<eta''quadRightarrowquad g(x)ne a .tag{3}$$
Put $eta:=min{eta', eta''}>0$. Then $(2)$ and $(3)$ together imply
$$0<|x-b|<etaquadRightarrowquad 0<|g(x)-a|<delta .tag{4}$$
Taking $(4)$ and letting $y:=g(x)$ in $(1)$ we finally obtain
$$0<|x-b|<etaquadRightarrowquad bigl|fbigl(g(x)bigr)-ellbigr|<epsilon .$$
answered Sep 4 '18 at 19:16
Christian BlatterChristian Blatter
173k7113326
$endgroup$
Your statement (ii) is correct: If $lim_{yto a} f(y)=ell$, $ lim_{xto b} g(x)=a$, and $g$ does not assume the value $a$ in a punctured neighborhood $dot U$ of $b$ then $lim_{xto b}fbigl(g(x)bigr)=ell$.
Proof. Let an $epsilon>0$ be given. Then there is a punctured neighborhood $dot V$ of $a$ such that $|f(y)-ell|<epsilon$ for all $yindot V$. Furthermore there is a punctured neighborhood $dot W$ of $b$ such that $g(x)in V$ for all $bindot W$. The set $dot Omega:=dot Wcapdot U$ is a punctured neighborhood of $b$ as well, and for all $xin dot Omega$ we have $g(x)ne a$, hence $g(x)in dot V$. This then implies that
$$bigl|fbigl(g(x)bigr)-ellbigr|<epsilonqquadforall xindotOmega ,$$
and proves the claim. Note that this proof also works when $a$ or $b$ are $ =pminfty$.
In $epsilon/delta$ terms exactly the same proof is much more cumbersome:
Let an $epsilon>0$ be given. Then there is a $delta>0$ such that
$$0<|y-a|<deltaquadRightarrowquad |f(y)-ell|<epsilon .tag{1}$$
Furthermore there is an $eta'>0$ such that
$$0<|x-b|<eta'quadRightarrowquad |g(x)-a|<delta .tag{2}$$
Since $g$ does not assume the value $a$ in a punctured neighborhood of $b$ there is an $eta''>0$ such that $$0<|x-b|<eta''quadRightarrowquad g(x)ne a .tag{3}$$
Put $eta:=min{eta', eta''}>0$. Then $(2)$ and $(3)$ together imply
$$0<|x-b|<etaquadRightarrowquad 0<|g(x)-a|<delta .tag{4}$$
Taking $(4)$ and letting $y:=g(x)$ in $(1)$ we finally obtain
$$0<|x-b|<etaquadRightarrowquad bigl|fbigl(g(x)bigr)-ellbigr|<epsilon .$$
answered Sep 4 '18 at 19:16
Christian BlatterChristian Blatter
173k7113326
edited Sep 6 '18 at 7:08
answered Sep 4 '18 at 19:16
Christian BlatterChristian Blatter
173k7113326
answered Sep 4 '18 at 19:16
Christian BlatterChristian Blatter
173k7113326
answered Sep 4 '18 at 19:16
Christian BlatterChristian Blatter
173k7113326
173k7113326
$begingroup$
Can you write your last statement (|f(g(x)-l| < $epsilon$ $forall$ x$in$ $Omega$) in the form of $epsilon$ $delta$ statement, because I am still confused if it satisfies the limit definition ($epsilon$ $delta$) or if that is too long just give the reason why this will be true for all $epsilon$>0 . Thanks for the answer
$endgroup$
– Steve
Sep 5 '18 at 11:05
$begingroup$
can we say that existence of the interval $Omega$ is always guaranteed because the existence of W is always guaranteed ? So,there always exists an interval of b such that |f(g(x))-l|<$epsilon$ , right?
$endgroup$
– Steve
Sep 5 '18 at 11:19
$begingroup$
Thanks for the edit. Now it is crystal clear.
$endgroup$
– Steve
Sep 6 '18 at 15:11
add a comment |
$begingroup$
Can you write your last statement (|f(g(x)-l| < $epsilon$ $forall$ x$in$ $Omega$) in the form of $epsilon$ $delta$ statement, because I am still confused if it satisfies the limit definition ($epsilon$ $delta$) or if that is too long just give the reason why this will be true for all $epsilon$>0 . Thanks for the answer
$endgroup$
– Steve
Sep 5 '18 at 11:05
$begingroup$
can we say that existence of the interval $Omega$ is always guaranteed because the existence of W is always guaranteed ? So,there always exists an interval of b such that |f(g(x))-l|<$epsilon$ , right?
$endgroup$
– Steve
Sep 5 '18 at 11:19
$begingroup$
Thanks for the edit. Now it is crystal clear.
$endgroup$
– Steve
Sep 6 '18 at 15:11
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Can you write your last statement (|f(g(x)-l| < $epsilon$ $forall$ x$in$ $Omega$) in the form of $epsilon$ $delta$ statement, because I am still confused if it satisfies the limit definition ($epsilon$ $delta$) or if that is too long just give the reason why this will be true for all $epsilon$>0 . Thanks for the answer
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– Steve
Sep 5 '18 at 11:05
$begingroup$
Can you write your last statement (|f(g(x)-l| < $epsilon$ $forall$ x$in$ $Omega$) in the form of $epsilon$ $delta$ statement, because I am still confused if it satisfies the limit definition ($epsilon$ $delta$) or if that is too long just give the reason why this will be true for all $epsilon$>0 . Thanks for the answer
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– Steve
Sep 5 '18 at 11:05
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can we say that existence of the interval $Omega$ is always guaranteed because the existence of W is always guaranteed ? So,there always exists an interval of b such that |f(g(x))-l|<$epsilon$ , right?
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– Steve
Sep 5 '18 at 11:19
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can we say that existence of the interval $Omega$ is always guaranteed because the existence of W is always guaranteed ? So,there always exists an interval of b such that |f(g(x))-l|<$epsilon$ , right?
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– Steve
Sep 5 '18 at 11:19
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Thanks for the edit. Now it is crystal clear.
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– Steve
Sep 6 '18 at 15:11
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Thanks for the edit. Now it is crystal clear.
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– Steve
Sep 6 '18 at 15:11
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$begingroup$
Your claim (ii) isn't correct in the first place: for all you know, $f$ can be discontinuous at $a$ and $g$ can be constantly equal to $a$, or be $g(x)=a+(x-b)sinfrac1{x-b}$.
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– Saucy O'Path
Sep 4 '18 at 17:27
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But that's what is exactly covered in (ii) "If f is discontinuous at a, then there exists an interval containing b such that g(x) $neq$ a for all x in I except possibly at b" which means that g(x) is not constantly equal to a in an interval around b.
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– Steve
Sep 4 '18 at 17:34