My attempt to show the conditions where if $lim_xto af(x)=bland lim_yto bg(y)=cimplieslim_xto a(gcirc f)(x)=c$
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I want to check if this way to prove the theorem about limits of compositions of functions is correct. Im equating logical statements but Im not sure if this formal proof is correct or I need to change some symbols.
We have some $f:Ato B$ and $g:Bto C$ where the composition of $(gcirc f)(x)$ is well defined. And we have that $lim_xto af(x)=b$ and $lim_yto bg(y)=c$ and we want to see what is needed for $lim_xto a(gcirc f)(x)=c$.
Expanding the definition of functional limits we have:
$$lim_xto af(x)=biff(forallvarepsilon>0,existsdelta>0:0<|x-a|<deltaimplies |f(x)-b|<varepsilon)$$
$$lim_yto bg(y)=ciff(forallgamma>0,existsvarepsilon>0:0<|y-b|<varepsilonimplies |g(y)-c|<gamma)$$
$$lim_xto a(gcirc f)(x)=ciff(forallgamma>0,existsdelta>0:0<|x-a|<deltaimplies |(gcirc f)(x)-c|<gamma)$$
Equating logical statements assuming that $f(x)=y$ we can see that all three only hold at once under two possible conditions:
If we change in the first statement $colorred<varepsilon$ by $colorgreen<varepsilon$ what means that exists a $delta$-neighborhood where $forall xin V_delta(a)setminusaimplies f(x)neq b$
If we change in the second statement $colorred<varepsilon$ by $colorgreen<varepsilon$ what means that $g(y)$ is continuous at $b$
real-analysis proof-verification
asked May 29 '16 at 3:08
Masacroso
11.6k41743
add a comment |Â
up vote
3
down vote
favorite
I want to check if this way to prove the theorem about limits of compositions of functions is correct. Im equating logical statements but Im not sure if this formal proof is correct or I need to change some symbols.
We have some $f:Ato B$ and $g:Bto C$ where the composition of $(gcirc f)(x)$ is well defined. And we have that $lim_xto af(x)=b$ and $lim_yto bg(y)=c$ and we want to see what is needed for $lim_xto a(gcirc f)(x)=c$.
Expanding the definition of functional limits we have:
$$lim_xto af(x)=biff(forallvarepsilon>0,existsdelta>0:0<|x-a|<deltaimplies |f(x)-b|<varepsilon)$$
$$lim_yto bg(y)=ciff(forallgamma>0,existsvarepsilon>0:0<|y-b|<varepsilonimplies |g(y)-c|<gamma)$$
$$lim_xto a(gcirc f)(x)=ciff(forallgamma>0,existsdelta>0:0<|x-a|<deltaimplies |(gcirc f)(x)-c|<gamma)$$
Equating logical statements assuming that $f(x)=y$ we can see that all three only hold at once under two possible conditions:
If we change in the first statement $colorred<varepsilon$ by $colorgreen<varepsilon$ what means that exists a $delta$-neighborhood where $forall xin V_delta(a)setminusaimplies f(x)neq b$
If we change in the second statement $colorred<varepsilon$ by $colorgreen<varepsilon$ what means that $g(y)$ is continuous at $b$
real-analysis proof-verification
asked May 29 '16 at 3:08
Masacroso
11.6k41743
3
Your observations are correct. And it is better to chose the first condition $f(x) neq b$ instead of choosing continuity of $g$.
– Paramanand Singh
May 29 '16 at 15:13
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I want to check if this way to prove the theorem about limits of compositions of functions is correct. Im equating logical statements but Im not sure if this formal proof is correct or I need to change some symbols.
We have some $f:Ato B$ and $g:Bto C$ where the composition of $(gcirc f)(x)$ is well defined. And we have that $lim_xto af(x)=b$ and $lim_yto bg(y)=c$ and we want to see what is needed for $lim_xto a(gcirc f)(x)=c$.
Expanding the definition of functional limits we have:
$$lim_xto af(x)=biff(forallvarepsilon>0,existsdelta>0:0<|x-a|<deltaimplies |f(x)-b|<varepsilon)$$
$$lim_yto bg(y)=ciff(forallgamma>0,existsvarepsilon>0:0<|y-b|<varepsilonimplies |g(y)-c|<gamma)$$
$$lim_xto a(gcirc f)(x)=ciff(forallgamma>0,existsdelta>0:0<|x-a|<deltaimplies |(gcirc f)(x)-c|<gamma)$$
Equating logical statements assuming that $f(x)=y$ we can see that all three only hold at once under two possible conditions:
If we change in the first statement $colorred<varepsilon$ by $colorgreen<varepsilon$ what means that exists a $delta$-neighborhood where $forall xin V_delta(a)setminusaimplies f(x)neq b$
If we change in the second statement $colorred<varepsilon$ by $colorgreen<varepsilon$ what means that $g(y)$ is continuous at $b$
real-analysis proof-verification
asked May 29 '16 at 3:08
Masacroso
11.6k41743
I want to check if this way to prove the theorem about limits of compositions of functions is correct. Im equating logical statements but Im not sure if this formal proof is correct or I need to change some symbols.
We have some $f:Ato B$ and $g:Bto C$ where the composition of $(gcirc f)(x)$ is well defined. And we have that $lim_xto af(x)=b$ and $lim_yto bg(y)=c$ and we want to see what is needed for $lim_xto a(gcirc f)(x)=c$.
Expanding the definition of functional limits we have:
$$lim_xto af(x)=biff(forallvarepsilon>0,existsdelta>0:0<|x-a|<deltaimplies |f(x)-b|<varepsilon)$$
$$lim_yto bg(y)=ciff(forallgamma>0,existsvarepsilon>0:0<|y-b|<varepsilonimplies |g(y)-c|<gamma)$$
$$lim_xto a(gcirc f)(x)=ciff(forallgamma>0,existsdelta>0:0<|x-a|<deltaimplies |(gcirc f)(x)-c|<gamma)$$
Equating logical statements assuming that $f(x)=y$ we can see that all three only hold at once under two possible conditions:
If we change in the first statement $colorred<varepsilon$ by $colorgreen<varepsilon$ what means that exists a $delta$-neighborhood where $forall xin V_delta(a)setminusaimplies f(x)neq b$
If we change in the second statement $colorred<varepsilon$ by $colorgreen<varepsilon$ what means that $g(y)$ is continuous at $b$
real-analysis proof-verification
real-analysis proof-verification
asked May 29 '16 at 3:08
Masacroso
11.6k41743
asked May 29 '16 at 3:08
Masacroso
11.6k41743
edited Sep 10 at 12:28
asked May 29 '16 at 3:08
Masacroso
11.6k41743
asked May 29 '16 at 3:08
Masacroso
11.6k41743
asked May 29 '16 at 3:08
Masacroso
11.6k41743
11.6k41743
3
Your observations are correct. And it is better to chose the first condition $f(x) neq b$ instead of choosing continuity of $g$.
– Paramanand Singh
May 29 '16 at 15:13
add a comment |Â
3
Your observations are correct. And it is better to chose the first condition $f(x) neq b$ instead of choosing continuity of $g$.
– Paramanand Singh
May 29 '16 at 15:13
3
3
Your observations are correct. And it is better to chose the first condition $f(x) neq b$ instead of choosing continuity of $g$.
– Paramanand Singh
May 29 '16 at 15:13
Your observations are correct. And it is better to chose the first condition $f(x) neq b$ instead of choosing continuity of $g$.
– Paramanand Singh
May 29 '16 at 15:13
add a comment |Â
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3
Your observations are correct. And it is better to chose the first condition $f(x) neq b$ instead of choosing continuity of $g$.
– Paramanand Singh
May 29 '16 at 15:13