Need Help : Proving polynomials are continuous, without circular reasoning












12












$begingroup$


I know there are a lot of answers regarding continuity of polynomials. But, this question is different.



We need to have $ lim_{xto a} {x^n} = a^n$ , $n in N$ , to be able to prove that polynomials are continuous. This fact is derived from the product rule (or may be it can't be, which is my question). The product rule is proved using square roots, so it assumes the existence of square roots. The fact that For every non negative number $x$, it's $n^{th}$ root exists, i.e. $x^n$ is invertible, assumes the continuity of $x^n$ because this is proven using Intermediate value Theorem.

Bam - Circular reasoning ! Or am I Wrong ?
Here's the only proof of product rule which I know :
Assume $ lim_{xto a} {f(x)} = L$ and $ lim_{xto a} {g(x)} = K$



Let $ϵ > 0$ be any positive number
Hence, $∃delta_1> 0 ∶ 0<|x-a|<δ_1⟹|f(x)-L|<sqrt{epsilon}$



And $∃δ_2>0 ∶ 0<|x-a|<δ_2 ⟹ |g(x)-K|<sqrt{epsilon}$



Let $δ=min{δ_1,δ_2}$



Hence, $0<|x-a|<δ$



⟹$|(f(x)-L)(g(x)-K)-0|<sqrt{epsilon} sqrt{epsilon} = ϵ$



Hence, $lim_{x to a} {(f(x)-L)(g(x)-K)} = 0$



⟹$lim_{x to a} {(f(x)g(x)-Kf(x)-Lg(x)+KL)} = 0$



And then the result follows.
Even if we use $epsilon$ in place of $√ϵ$ , we end up with
$0<|x-a|<δ⟹|(f(x)-L)(g(x)-K)-0|<epsiloncdot epsilon = ϵ^2$ , and then we have to prove that the range of $epsilon^2$ is $[0,infty]$, which amounts to proving that for each number in $[0,infty]$ , a corresponding square root exists.






real-analysis limits continuity epsilon-delta







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $lim_{xto a}$ and $Lim_{xto a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit.
    $endgroup$
    – Martin Sleziak
    Jan 7 at 8:57












  • $begingroup$
    Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
    $endgroup$
    – Steve
    Jan 7 at 9:01








  • 2




    $begingroup$
    Why don't you use the fact that $$|f(x) g(x) - LK|leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $epsilon /2$?
    $endgroup$
    – Paramanand Singh
    Jan 7 at 9:10






  • 1




    $begingroup$
    Use the fact that $g(x) to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<epsilon/(2(|L|+1))$ to handle second term.
    $endgroup$
    – Paramanand Singh
    Jan 7 at 9:29






  • 2




    $begingroup$
    In general when you are dealing with $epsilon, delta$ proofs the expression based on these $epsilon, delta$ should not involve operations other than $+, -, times, /$.
    $endgroup$
    – Paramanand Singh
    Jan 7 at 9:32
















12












$begingroup$


I know there are a lot of answers regarding continuity of polynomials. But, this question is different.



We need to have $ lim_{xto a} {x^n} = a^n$ , $n in N$ , to be able to prove that polynomials are continuous. This fact is derived from the product rule (or may be it can't be, which is my question). The product rule is proved using square roots, so it assumes the existence of square roots. The fact that For every non negative number $x$, it's $n^{th}$ root exists, i.e. $x^n$ is invertible, assumes the continuity of $x^n$ because this is proven using Intermediate value Theorem.

Bam - Circular reasoning ! Or am I Wrong ?
Here's the only proof of product rule which I know :
Assume $ lim_{xto a} {f(x)} = L$ and $ lim_{xto a} {g(x)} = K$



Let $ϵ > 0$ be any positive number
Hence, $∃delta_1> 0 ∶ 0<|x-a|<δ_1⟹|f(x)-L|<sqrt{epsilon}$



And $∃δ_2>0 ∶ 0<|x-a|<δ_2 ⟹ |g(x)-K|<sqrt{epsilon}$



Let $δ=min{δ_1,δ_2}$



Hence, $0<|x-a|<δ$



⟹$|(f(x)-L)(g(x)-K)-0|<sqrt{epsilon} sqrt{epsilon} = ϵ$



Hence, $lim_{x to a} {(f(x)-L)(g(x)-K)} = 0$



⟹$lim_{x to a} {(f(x)g(x)-Kf(x)-Lg(x)+KL)} = 0$



And then the result follows.
Even if we use $epsilon$ in place of $√ϵ$ , we end up with
$0<|x-a|<δ⟹|(f(x)-L)(g(x)-K)-0|<epsiloncdot epsilon = ϵ^2$ , and then we have to prove that the range of $epsilon^2$ is $[0,infty]$, which amounts to proving that for each number in $[0,infty]$ , a corresponding square root exists.






real-analysis limits continuity epsilon-delta







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $lim_{xto a}$ and $Lim_{xto a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit.
    $endgroup$
    – Martin Sleziak
    Jan 7 at 8:57












  • $begingroup$
    Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
    $endgroup$
    – Steve
    Jan 7 at 9:01








  • 2




    $begingroup$
    Why don't you use the fact that $$|f(x) g(x) - LK|leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $epsilon /2$?
    $endgroup$
    – Paramanand Singh
    Jan 7 at 9:10






  • 1




    $begingroup$
    Use the fact that $g(x) to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<epsilon/(2(|L|+1))$ to handle second term.
    $endgroup$
    – Paramanand Singh
    Jan 7 at 9:29






  • 2




    $begingroup$
    In general when you are dealing with $epsilon, delta$ proofs the expression based on these $epsilon, delta$ should not involve operations other than $+, -, times, /$.
    $endgroup$
    – Paramanand Singh
    Jan 7 at 9:32














12












12








12


0



$begingroup$


I know there are a lot of answers regarding continuity of polynomials. But, this question is different.



We need to have $ lim_{xto a} {x^n} = a^n$ , $n in N$ , to be able to prove that polynomials are continuous. This fact is derived from the product rule (or may be it can't be, which is my question). The product rule is proved using square roots, so it assumes the existence of square roots. The fact that For every non negative number $x$, it's $n^{th}$ root exists, i.e. $x^n$ is invertible, assumes the continuity of $x^n$ because this is proven using Intermediate value Theorem.

Bam - Circular reasoning ! Or am I Wrong ?
Here's the only proof of product rule which I know :
Assume $ lim_{xto a} {f(x)} = L$ and $ lim_{xto a} {g(x)} = K$



Let $ϵ > 0$ be any positive number
Hence, $∃delta_1> 0 ∶ 0<|x-a|<δ_1⟹|f(x)-L|<sqrt{epsilon}$



And $∃δ_2>0 ∶ 0<|x-a|<δ_2 ⟹ |g(x)-K|<sqrt{epsilon}$



Let $δ=min{δ_1,δ_2}$



Hence, $0<|x-a|<δ$



⟹$|(f(x)-L)(g(x)-K)-0|<sqrt{epsilon} sqrt{epsilon} = ϵ$



Hence, $lim_{x to a} {(f(x)-L)(g(x)-K)} = 0$



⟹$lim_{x to a} {(f(x)g(x)-Kf(x)-Lg(x)+KL)} = 0$



And then the result follows.
Even if we use $epsilon$ in place of $√ϵ$ , we end up with
$0<|x-a|<δ⟹|(f(x)-L)(g(x)-K)-0|<epsiloncdot epsilon = ϵ^2$ , and then we have to prove that the range of $epsilon^2$ is $[0,infty]$, which amounts to proving that for each number in $[0,infty]$ , a corresponding square root exists.






real-analysis limits continuity epsilon-delta







share|cite|improve this question











$endgroup$




I know there are a lot of answers regarding continuity of polynomials. But, this question is different.



We need to have $ lim_{xto a} {x^n} = a^n$ , $n in N$ , to be able to prove that polynomials are continuous. This fact is derived from the product rule (or may be it can't be, which is my question). The product rule is proved using square roots, so it assumes the existence of square roots. The fact that For every non negative number $x$, it's $n^{th}$ root exists, i.e. $x^n$ is invertible, assumes the continuity of $x^n$ because this is proven using Intermediate value Theorem.

Bam - Circular reasoning ! Or am I Wrong ?
Here's the only proof of product rule which I know :
Assume $ lim_{xto a} {f(x)} = L$ and $ lim_{xto a} {g(x)} = K$



Let $ϵ > 0$ be any positive number
Hence, $∃delta_1> 0 ∶ 0<|x-a|<δ_1⟹|f(x)-L|<sqrt{epsilon}$



And $∃δ_2>0 ∶ 0<|x-a|<δ_2 ⟹ |g(x)-K|<sqrt{epsilon}$



Let $δ=min{δ_1,δ_2}$



Hence, $0<|x-a|<δ$



⟹$|(f(x)-L)(g(x)-K)-0|<sqrt{epsilon} sqrt{epsilon} = ϵ$



Hence, $lim_{x to a} {(f(x)-L)(g(x)-K)} = 0$



⟹$lim_{x to a} {(f(x)g(x)-Kf(x)-Lg(x)+KL)} = 0$



And then the result follows.
Even if we use $epsilon$ in place of $√ϵ$ , we end up with
$0<|x-a|<δ⟹|(f(x)-L)(g(x)-K)-0|<epsiloncdot epsilon = ϵ^2$ , and then we have to prove that the range of $epsilon^2$ is $[0,infty]$, which amounts to proving that for each number in $[0,infty]$ , a corresponding square root exists.






real-analysis limits continuity epsilon-delta




real-analysis limits continuity epsilon-delta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 9:09









Martin Sleziak

44.9k10122276




44.9k10122276










asked Jan 7 at 8:39









SteveSteve

828




828








  • 1




    $begingroup$
    You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $lim_{xto a}$ and $Lim_{xto a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit.
    $endgroup$
    – Martin Sleziak
    Jan 7 at 8:57












  • $begingroup$
    Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
    $endgroup$
    – Steve
    Jan 7 at 9:01








  • 2




    $begingroup$
    Why don't you use the fact that $$|f(x) g(x) - LK|leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $epsilon /2$?
    $endgroup$
    – Paramanand Singh
    Jan 7 at 9:10






  • 1




    $begingroup$
    Use the fact that $g(x) to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<epsilon/(2(|L|+1))$ to handle second term.
    $endgroup$
    – Paramanand Singh
    Jan 7 at 9:29






  • 2




    $begingroup$
    In general when you are dealing with $epsilon, delta$ proofs the expression based on these $epsilon, delta$ should not involve operations other than $+, -, times, /$.
    $endgroup$
    – Paramanand Singh
    Jan 7 at 9:32














  • 1




    $begingroup$
    You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $lim_{xto a}$ and $Lim_{xto a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit.
    $endgroup$
    – Martin Sleziak
    Jan 7 at 8:57












  • $begingroup$
    Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
    $endgroup$
    – Steve
    Jan 7 at 9:01








  • 2




    $begingroup$
    Why don't you use the fact that $$|f(x) g(x) - LK|leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $epsilon /2$?
    $endgroup$
    – Paramanand Singh
    Jan 7 at 9:10






  • 1




    $begingroup$
    Use the fact that $g(x) to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<epsilon/(2(|L|+1))$ to handle second term.
    $endgroup$
    – Paramanand Singh
    Jan 7 at 9:29






  • 2




    $begingroup$
    In general when you are dealing with $epsilon, delta$ proofs the expression based on these $epsilon, delta$ should not involve operations other than $+, -, times, /$.
    $endgroup$
    – Paramanand Singh
    Jan 7 at 9:32








1




1




$begingroup$
You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $lim_{xto a}$ and $Lim_{xto a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit.
$endgroup$
– Martin Sleziak
Jan 7 at 8:57






$begingroup$
You can use MathJax in a single formula - you don't have to write $f(x)$<$g(x)$ and similar formulas. Also I see two different notations $lim_{xto a}$ and $Lim_{xto a}$ in the post - but without an explanation whether they are intended to represent two different things or they are both just different notations for the usual limit.
$endgroup$
– Martin Sleziak
Jan 7 at 8:57














$begingroup$
Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
$endgroup$
– Steve
Jan 7 at 9:01






$begingroup$
Thank you. I have edited the question. The limit with lowercase L and uppercase L are not different, it was my mistake.
$endgroup$
– Steve
Jan 7 at 9:01






2




2




$begingroup$
Why don't you use the fact that $$|f(x) g(x) - LK|leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $epsilon /2$?
$endgroup$
– Paramanand Singh
Jan 7 at 9:10




$begingroup$
Why don't you use the fact that $$|f(x) g(x) - LK|leq |g(x) ||f(x) - L|+|L||g(x) - K|$$ and show that each term on right can be made less than $epsilon /2$?
$endgroup$
– Paramanand Singh
Jan 7 at 9:10




1




1




$begingroup$
Use the fact that $g(x) to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<epsilon/(2(|L|+1))$ to handle second term.
$endgroup$
– Paramanand Singh
Jan 7 at 9:29




$begingroup$
Use the fact that $g(x) to K$ and hence $|g(x) |<|K|+1$ and we can take $|f(x) - L|<epsilon /(2(|K|+1))$. This handles first term. In similar manner you can take $|g(x) - K|<epsilon/(2(|L|+1))$ to handle second term.
$endgroup$
– Paramanand Singh
Jan 7 at 9:29




2




2




$begingroup$
In general when you are dealing with $epsilon, delta$ proofs the expression based on these $epsilon, delta$ should not involve operations other than $+, -, times, /$.
$endgroup$
– Paramanand Singh
Jan 7 at 9:32




$begingroup$
In general when you are dealing with $epsilon, delta$ proofs the expression based on these $epsilon, delta$ should not involve operations other than $+, -, times, /$.
$endgroup$
– Paramanand Singh
Jan 7 at 9:32










2 Answers
2






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17












$begingroup$

The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that $0 < epsilon < 1$, in which case $epsilon^2 < epsilon$. Or you could use a convergent sequence $x_n$ such that $displaystyle lim_{n mathop to infty} x_n = a$, and show that



$$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$






share|cite|improve this answer











$endgroup$





















    10












    $begingroup$

    You don't need existence of square root. You just need to have that $forall epsilon > 0 exists epsilon^prime > 0$ s. t. ${epsilon^prime}^2 < epsilon$. Then, you'll have $|(f(x) - K)(g(x)-L)| < {epsilon^prime}^2 < epsilon$ which gives you what you need anyway. You don't need the bound $epsilon$ to actually be reachable.



    For $epsilon < 1$, $epsilon$ itself can serve as $epsilon^prime$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
      $endgroup$
      – leftaroundabout
      Jan 7 at 20:59













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    17












    $begingroup$

    The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that $0 < epsilon < 1$, in which case $epsilon^2 < epsilon$. Or you could use a convergent sequence $x_n$ such that $displaystyle lim_{n mathop to infty} x_n = a$, and show that



    $$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$






    share|cite|improve this answer











    $endgroup$


















      17












      $begingroup$

      The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that $0 < epsilon < 1$, in which case $epsilon^2 < epsilon$. Or you could use a convergent sequence $x_n$ such that $displaystyle lim_{n mathop to infty} x_n = a$, and show that



      $$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$






      share|cite|improve this answer











      $endgroup$
















        17












        17








        17





        $begingroup$

        The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that $0 < epsilon < 1$, in which case $epsilon^2 < epsilon$. Or you could use a convergent sequence $x_n$ such that $displaystyle lim_{n mathop to infty} x_n = a$, and show that



        $$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$






        share|cite|improve this answer











        $endgroup$



        The proof that the product of continuous functions is also continuous does not need to use square roots. You could add the condition that $0 < epsilon < 1$, in which case $epsilon^2 < epsilon$. Or you could use a convergent sequence $x_n$ such that $displaystyle lim_{n mathop to infty} x_n = a$, and show that



        $$lim_{n to infty}f(x_n)g(x_n) = fleft( lim_{n to infty} x_n right)gleft(lim_{n to infty} x_nright) = f(a)g(a)$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 7 at 11:12









        Mutantoe

        623513




        623513










        answered Jan 7 at 9:14









        gandalf61gandalf61

        9,164825




        9,164825























            10












            $begingroup$

            You don't need existence of square root. You just need to have that $forall epsilon > 0 exists epsilon^prime > 0$ s. t. ${epsilon^prime}^2 < epsilon$. Then, you'll have $|(f(x) - K)(g(x)-L)| < {epsilon^prime}^2 < epsilon$ which gives you what you need anyway. You don't need the bound $epsilon$ to actually be reachable.



            For $epsilon < 1$, $epsilon$ itself can serve as $epsilon^prime$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
              $endgroup$
              – leftaroundabout
              Jan 7 at 20:59


















            10












            $begingroup$

            You don't need existence of square root. You just need to have that $forall epsilon > 0 exists epsilon^prime > 0$ s. t. ${epsilon^prime}^2 < epsilon$. Then, you'll have $|(f(x) - K)(g(x)-L)| < {epsilon^prime}^2 < epsilon$ which gives you what you need anyway. You don't need the bound $epsilon$ to actually be reachable.



            For $epsilon < 1$, $epsilon$ itself can serve as $epsilon^prime$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
              $endgroup$
              – leftaroundabout
              Jan 7 at 20:59
















            10












            10








            10





            $begingroup$

            You don't need existence of square root. You just need to have that $forall epsilon > 0 exists epsilon^prime > 0$ s. t. ${epsilon^prime}^2 < epsilon$. Then, you'll have $|(f(x) - K)(g(x)-L)| < {epsilon^prime}^2 < epsilon$ which gives you what you need anyway. You don't need the bound $epsilon$ to actually be reachable.



            For $epsilon < 1$, $epsilon$ itself can serve as $epsilon^prime$.






            share|cite|improve this answer









            $endgroup$



            You don't need existence of square root. You just need to have that $forall epsilon > 0 exists epsilon^prime > 0$ s. t. ${epsilon^prime}^2 < epsilon$. Then, you'll have $|(f(x) - K)(g(x)-L)| < {epsilon^prime}^2 < epsilon$ which gives you what you need anyway. You don't need the bound $epsilon$ to actually be reachable.



            For $epsilon < 1$, $epsilon$ itself can serve as $epsilon^prime$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 7 at 9:09









            Todor MarkovTodor Markov

            2,420412




            2,420412












            • $begingroup$
              IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
              $endgroup$
              – leftaroundabout
              Jan 7 at 20:59




















            • $begingroup$
              IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
              $endgroup$
              – leftaroundabout
              Jan 7 at 20:59


















            $begingroup$
            IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
            $endgroup$
            – leftaroundabout
            Jan 7 at 20:59






            $begingroup$
            IOW, you don't need the fact that $xmapsto x^2$ is continuous, just that it's monotonic for $x>0$.
            $endgroup$
            – leftaroundabout
            Jan 7 at 20:59




















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