Statement and proof of the sequential characterization of limits












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So frequently on this site I will use the sequential characterizations of limits and continuity in my answers only to find the OP is unfamiliar with it. It is such a crucial notion that I am going to write the statement and proof here so that I (and others) can link it in answers.






real-analysis limits







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    2












    $begingroup$


    So frequently on this site I will use the sequential characterizations of limits and continuity in my answers only to find the OP is unfamiliar with it. It is such a crucial notion that I am going to write the statement and proof here so that I (and others) can link it in answers.






    real-analysis limits







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      So frequently on this site I will use the sequential characterizations of limits and continuity in my answers only to find the OP is unfamiliar with it. It is such a crucial notion that I am going to write the statement and proof here so that I (and others) can link it in answers.






      real-analysis limits







      share|cite|improve this question









      $endgroup$




      So frequently on this site I will use the sequential characterizations of limits and continuity in my answers only to find the OP is unfamiliar with it. It is such a crucial notion that I am going to write the statement and proof here so that I (and others) can link it in answers.






      real-analysis limits




      real-analysis limits






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      asked Feb 23 '18 at 21:23









      David ReedDavid Reed

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          $begingroup$

          Theorem (Sequential Characterization of Limits):



          Let $I$ be an open interval, $a in I$, and $f:I to mathbb{R}$, then:



          $\$



          $$lim_{x to a}f(x) = L text{ if and only if }lim_{n to infty}f(x_n) = L text{ for every sequence }x_n in I - { a } text { satisfying } \ x_n to a text{ as } n to infty$$



          $\$



          Proof:



          $\$



          $Rightarrow:$



          Suppose $lim_{x to a}f(x) = L$ and let ${x_n}$ be a sequence such that $x_n to a$. Let $varepsilon > 0$ be given. Choose $delta > 0$ such that $$ 0 < vert x-avert < delta implies vert f(x) - L vert < varepsilon$$



          Since $delta > 0$ and $x_n to a$ we can choose an $N in mathbb{N}$ such that $$ n geq N implies vert x_n - a vert < delta$$



          Combining both statements we have $$n geq N implies |f(x_n) - L| < varepsilon$$



          and hence $lim_{n to infty}f(x_n) = L$



          $Leftarrow$:



          By contradiction, suppose $x_n to a$ implies $f(x_n) to L$, but $ lim_{x to a}f(x) neq L$, then $$left(exists , varepsilon_0 > 0right)left(forall delta > 0right)left(exists x in I right)left(0 < vert x - a vert < delta text{ but } vert f(x) - L vert geq varepsilon_0right) $$



          $\$



          Thus we can define a sequence ${x_n}$ as follows: For each $delta = 1/n$, choose $x_n in I - {a}$ such that: $$ 0 < vert x_n - a vert < 1/n text{ and } vert f(x_n) - L vert geqvarepsilon_0$$



          By the squeeze theorem, $x_n to a$ and thus by assumption $f(x_n) to L$. However, by construction, $vert f(x_n) - L vert geq varepsilon_0$ for all $n$ and thus ${f(x_n)}$ cannot converge to $L$. We have arrived at a contradiction. This completes the proof.






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            $begingroup$

            Theorem (Sequential Characterization of Limits):



            Let $I$ be an open interval, $a in I$, and $f:I to mathbb{R}$, then:



            $\$



            $$lim_{x to a}f(x) = L text{ if and only if }lim_{n to infty}f(x_n) = L text{ for every sequence }x_n in I - { a } text { satisfying } \ x_n to a text{ as } n to infty$$



            $\$



            Proof:



            $\$



            $Rightarrow:$



            Suppose $lim_{x to a}f(x) = L$ and let ${x_n}$ be a sequence such that $x_n to a$. Let $varepsilon > 0$ be given. Choose $delta > 0$ such that $$ 0 < vert x-avert < delta implies vert f(x) - L vert < varepsilon$$



            Since $delta > 0$ and $x_n to a$ we can choose an $N in mathbb{N}$ such that $$ n geq N implies vert x_n - a vert < delta$$



            Combining both statements we have $$n geq N implies |f(x_n) - L| < varepsilon$$



            and hence $lim_{n to infty}f(x_n) = L$



            $Leftarrow$:



            By contradiction, suppose $x_n to a$ implies $f(x_n) to L$, but $ lim_{x to a}f(x) neq L$, then $$left(exists , varepsilon_0 > 0right)left(forall delta > 0right)left(exists x in I right)left(0 < vert x - a vert < delta text{ but } vert f(x) - L vert geq varepsilon_0right) $$



            $\$



            Thus we can define a sequence ${x_n}$ as follows: For each $delta = 1/n$, choose $x_n in I - {a}$ such that: $$ 0 < vert x_n - a vert < 1/n text{ and } vert f(x_n) - L vert geqvarepsilon_0$$



            By the squeeze theorem, $x_n to a$ and thus by assumption $f(x_n) to L$. However, by construction, $vert f(x_n) - L vert geq varepsilon_0$ for all $n$ and thus ${f(x_n)}$ cannot converge to $L$. We have arrived at a contradiction. This completes the proof.






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              Theorem (Sequential Characterization of Limits):



              Let $I$ be an open interval, $a in I$, and $f:I to mathbb{R}$, then:



              $\$



              $$lim_{x to a}f(x) = L text{ if and only if }lim_{n to infty}f(x_n) = L text{ for every sequence }x_n in I - { a } text { satisfying } \ x_n to a text{ as } n to infty$$



              $\$



              Proof:



              $\$



              $Rightarrow:$



              Suppose $lim_{x to a}f(x) = L$ and let ${x_n}$ be a sequence such that $x_n to a$. Let $varepsilon > 0$ be given. Choose $delta > 0$ such that $$ 0 < vert x-avert < delta implies vert f(x) - L vert < varepsilon$$



              Since $delta > 0$ and $x_n to a$ we can choose an $N in mathbb{N}$ such that $$ n geq N implies vert x_n - a vert < delta$$



              Combining both statements we have $$n geq N implies |f(x_n) - L| < varepsilon$$



              and hence $lim_{n to infty}f(x_n) = L$



              $Leftarrow$:



              By contradiction, suppose $x_n to a$ implies $f(x_n) to L$, but $ lim_{x to a}f(x) neq L$, then $$left(exists , varepsilon_0 > 0right)left(forall delta > 0right)left(exists x in I right)left(0 < vert x - a vert < delta text{ but } vert f(x) - L vert geq varepsilon_0right) $$



              $\$



              Thus we can define a sequence ${x_n}$ as follows: For each $delta = 1/n$, choose $x_n in I - {a}$ such that: $$ 0 < vert x_n - a vert < 1/n text{ and } vert f(x_n) - L vert geqvarepsilon_0$$



              By the squeeze theorem, $x_n to a$ and thus by assumption $f(x_n) to L$. However, by construction, $vert f(x_n) - L vert geq varepsilon_0$ for all $n$ and thus ${f(x_n)}$ cannot converge to $L$. We have arrived at a contradiction. This completes the proof.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                Theorem (Sequential Characterization of Limits):



                Let $I$ be an open interval, $a in I$, and $f:I to mathbb{R}$, then:



                $\$



                $$lim_{x to a}f(x) = L text{ if and only if }lim_{n to infty}f(x_n) = L text{ for every sequence }x_n in I - { a } text { satisfying } \ x_n to a text{ as } n to infty$$



                $\$



                Proof:



                $\$



                $Rightarrow:$



                Suppose $lim_{x to a}f(x) = L$ and let ${x_n}$ be a sequence such that $x_n to a$. Let $varepsilon > 0$ be given. Choose $delta > 0$ such that $$ 0 < vert x-avert < delta implies vert f(x) - L vert < varepsilon$$



                Since $delta > 0$ and $x_n to a$ we can choose an $N in mathbb{N}$ such that $$ n geq N implies vert x_n - a vert < delta$$



                Combining both statements we have $$n geq N implies |f(x_n) - L| < varepsilon$$



                and hence $lim_{n to infty}f(x_n) = L$



                $Leftarrow$:



                By contradiction, suppose $x_n to a$ implies $f(x_n) to L$, but $ lim_{x to a}f(x) neq L$, then $$left(exists , varepsilon_0 > 0right)left(forall delta > 0right)left(exists x in I right)left(0 < vert x - a vert < delta text{ but } vert f(x) - L vert geq varepsilon_0right) $$



                $\$



                Thus we can define a sequence ${x_n}$ as follows: For each $delta = 1/n$, choose $x_n in I - {a}$ such that: $$ 0 < vert x_n - a vert < 1/n text{ and } vert f(x_n) - L vert geqvarepsilon_0$$



                By the squeeze theorem, $x_n to a$ and thus by assumption $f(x_n) to L$. However, by construction, $vert f(x_n) - L vert geq varepsilon_0$ for all $n$ and thus ${f(x_n)}$ cannot converge to $L$. We have arrived at a contradiction. This completes the proof.






                share|cite|improve this answer











                $endgroup$



                Theorem (Sequential Characterization of Limits):



                Let $I$ be an open interval, $a in I$, and $f:I to mathbb{R}$, then:



                $\$



                $$lim_{x to a}f(x) = L text{ if and only if }lim_{n to infty}f(x_n) = L text{ for every sequence }x_n in I - { a } text { satisfying } \ x_n to a text{ as } n to infty$$



                $\$



                Proof:



                $\$



                $Rightarrow:$



                Suppose $lim_{x to a}f(x) = L$ and let ${x_n}$ be a sequence such that $x_n to a$. Let $varepsilon > 0$ be given. Choose $delta > 0$ such that $$ 0 < vert x-avert < delta implies vert f(x) - L vert < varepsilon$$



                Since $delta > 0$ and $x_n to a$ we can choose an $N in mathbb{N}$ such that $$ n geq N implies vert x_n - a vert < delta$$



                Combining both statements we have $$n geq N implies |f(x_n) - L| < varepsilon$$



                and hence $lim_{n to infty}f(x_n) = L$



                $Leftarrow$:



                By contradiction, suppose $x_n to a$ implies $f(x_n) to L$, but $ lim_{x to a}f(x) neq L$, then $$left(exists , varepsilon_0 > 0right)left(forall delta > 0right)left(exists x in I right)left(0 < vert x - a vert < delta text{ but } vert f(x) - L vert geq varepsilon_0right) $$



                $\$



                Thus we can define a sequence ${x_n}$ as follows: For each $delta = 1/n$, choose $x_n in I - {a}$ such that: $$ 0 < vert x_n - a vert < 1/n text{ and } vert f(x_n) - L vert geqvarepsilon_0$$



                By the squeeze theorem, $x_n to a$ and thus by assumption $f(x_n) to L$. However, by construction, $vert f(x_n) - L vert geq varepsilon_0$ for all $n$ and thus ${f(x_n)}$ cannot converge to $L$. We have arrived at a contradiction. This completes the proof.







                share|cite|improve this answer














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                edited Jan 21 at 2:50

























                answered Feb 23 '18 at 21:23









                David ReedDavid Reed

                2,3191422




                2,3191422






























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