How to prove that $lim_{xrightarrow0}frac{1}{2x+1}=1$ using $delta-epsilon$ definition?












0












$begingroup$


How to prove that $limlimits_{xrightarrow 0}frac{1}{2x+1}=1$ using $delta-epsilon$ definition?



I have the following so far-



$|x-0|<delta$ and $left|frac{1}{2x+1}-1right|<epsilon$



which boils to $left|frac{-2x}{2x+1}right|<epsilon$



I'm trying to figure out a way to factor out an $x$ but don't know how






calculus limits proof-verification proof-writing epsilon-delta







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @obscurans Thanks for you effort to edit posts into better shape. I'll just mention that you can use lim to get nicer rendering and also proper spacing for the limit symbol. For example $lim_{ntoinfty} x_n$ $lim_{ntoinfty} x_n$ or $limlimits_{ntoinfty} x_n$ $limlimits_{ntoinfty} x_n$.
    $endgroup$
    – Martin Sleziak
    Dec 19 '18 at 5:41
















0












$begingroup$


How to prove that $limlimits_{xrightarrow 0}frac{1}{2x+1}=1$ using $delta-epsilon$ definition?



I have the following so far-



$|x-0|<delta$ and $left|frac{1}{2x+1}-1right|<epsilon$



which boils to $left|frac{-2x}{2x+1}right|<epsilon$



I'm trying to figure out a way to factor out an $x$ but don't know how






calculus limits proof-verification proof-writing epsilon-delta







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @obscurans Thanks for you effort to edit posts into better shape. I'll just mention that you can use lim to get nicer rendering and also proper spacing for the limit symbol. For example $lim_{ntoinfty} x_n$ $lim_{ntoinfty} x_n$ or $limlimits_{ntoinfty} x_n$ $limlimits_{ntoinfty} x_n$.
    $endgroup$
    – Martin Sleziak
    Dec 19 '18 at 5:41














0












0








0


3



$begingroup$


How to prove that $limlimits_{xrightarrow 0}frac{1}{2x+1}=1$ using $delta-epsilon$ definition?



I have the following so far-



$|x-0|<delta$ and $left|frac{1}{2x+1}-1right|<epsilon$



which boils to $left|frac{-2x}{2x+1}right|<epsilon$



I'm trying to figure out a way to factor out an $x$ but don't know how






calculus limits proof-verification proof-writing epsilon-delta







share|cite|improve this question











$endgroup$




How to prove that $limlimits_{xrightarrow 0}frac{1}{2x+1}=1$ using $delta-epsilon$ definition?



I have the following so far-



$|x-0|<delta$ and $left|frac{1}{2x+1}-1right|<epsilon$



which boils to $left|frac{-2x}{2x+1}right|<epsilon$



I'm trying to figure out a way to factor out an $x$ but don't know how






calculus limits proof-verification proof-writing epsilon-delta




calculus limits proof-verification proof-writing epsilon-delta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 5:39









Martin Sleziak

44.7k10119272




44.7k10119272










asked Dec 19 '18 at 4:26









John RawlsJohn Rawls

1,253619




1,253619








  • 1




    $begingroup$
    @obscurans Thanks for you effort to edit posts into better shape. I'll just mention that you can use lim to get nicer rendering and also proper spacing for the limit symbol. For example $lim_{ntoinfty} x_n$ $lim_{ntoinfty} x_n$ or $limlimits_{ntoinfty} x_n$ $limlimits_{ntoinfty} x_n$.
    $endgroup$
    – Martin Sleziak
    Dec 19 '18 at 5:41














  • 1




    $begingroup$
    @obscurans Thanks for you effort to edit posts into better shape. I'll just mention that you can use lim to get nicer rendering and also proper spacing for the limit symbol. For example $lim_{ntoinfty} x_n$ $lim_{ntoinfty} x_n$ or $limlimits_{ntoinfty} x_n$ $limlimits_{ntoinfty} x_n$.
    $endgroup$
    – Martin Sleziak
    Dec 19 '18 at 5:41








1




1




$begingroup$
@obscurans Thanks for you effort to edit posts into better shape. I'll just mention that you can use lim to get nicer rendering and also proper spacing for the limit symbol. For example $lim_{ntoinfty} x_n$ $lim_{ntoinfty} x_n$ or $limlimits_{ntoinfty} x_n$ $limlimits_{ntoinfty} x_n$.
$endgroup$
– Martin Sleziak
Dec 19 '18 at 5:41




$begingroup$
@obscurans Thanks for you effort to edit posts into better shape. I'll just mention that you can use lim to get nicer rendering and also proper spacing for the limit symbol. For example $lim_{ntoinfty} x_n$ $lim_{ntoinfty} x_n$ or $limlimits_{ntoinfty} x_n$ $limlimits_{ntoinfty} x_n$.
$endgroup$
– Martin Sleziak
Dec 19 '18 at 5:41










3 Answers
3






active

oldest

votes


















2












$begingroup$

So you want
$|dfrac{2x}{2x+1}|<epsilon
$.



There are two ways go here.



For the first,
note that you don't need
the best bounds,
just one that will work.



Soppose
$|x| < dfrac14$.
Then
$|2x| < dfrac12$
so
$|1+2x| > 1-2cdot dfrac14
= dfrac12$.
Therefore
$|dfrac{2x}{2x+1}|
lt |dfrac{2x}{frac12}|
=|4x|
$.
To make
$|4x| < epsilon$,
you just need
$|x| < dfrac{epsilon}{4}$.



Therefore,
if $|x| < min(dfrac14, dfrac{epsilon}{4})$,
we have
$|dfrac{2x}{2x+1}|<epsilon
$.



If you want the best bounds,
$|dfrac{2x}{2x+1}|<epsilon
$
is the same as
$dfrac1{epsilon}
lt |dfrac{2x+1}{2x}|
= |1+dfrac{1}{2x}|
$.



If $x > 0$,
$|1+dfrac{1}{2x}|
=1+dfrac{1}{2x}
$
so we want
$dfrac1{epsilon}
lt 1+dfrac{1}{2x}
$
or
$dfrac1{epsilon}-1
lt dfrac{1}{2x}
$
or,
assuming $epsilon < 1$,
$2x
lt dfrac1{dfrac1{epsilon}-1}
= dfrac{epsilon}{1-epsilon}
$
or
$x
ltdfrac{epsilon}{2(1-epsilon)}
$.



If $x < 0$,
to make
$dfrac1{epsilon}
lt |1+dfrac{1}{2x}|
$
be easy to work with,
we need
$1+dfrac{1}{2x} < 0$
or
$x > -dfrac12$.



If this holds,
then
$|1+dfrac{1}{2x}|
=-dfrac{1}{2x}-1
$,
so we want
$-dfrac{1}{2x}-1
gt dfrac1{epsilon}
$
or
$-dfrac{1}{2x}
gt dfrac1{epsilon}+1
$
or
$-2x
lt dfrac1{dfrac1{epsilon}+1}
$
or
$-x
lt dfrac{epsilon}{2(1+epsilon)}
$.



I always prefer to get
a simple bound
which is not the best.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $midfrac{-2x}{2x+1}mid=midfrac{-2}{2+frac1x}midltmidfrac2{frac 1x}midltepsilon$. So make $mid xmidltfrac{epsilon}2$. That is, $delta =frac {epsilon}2$.






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      For $quaddelta<dfrac{varepsilon}{2(1-varepsilon)}impliesquaddisplaystyleleft|dfrac{-2x}{2x+1}right|<dfrac{2delta}{2delta+1}=dfrac{1}{1+frac{1}{2delta}}<varepsilon$






      share|cite|improve this answer











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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        So you want
        $|dfrac{2x}{2x+1}|<epsilon
        $        .



        There are two ways go here.



        For the first,
        note that you don't need
        the best bounds,
        just one that will work.



        Soppose
        $|x| < dfrac14$.
        Then
        $|2x| < dfrac12$
        so
        $|1+2x| > 1-2cdot dfrac14
        = dfrac12$        .
        Therefore
        $|dfrac{2x}{2x+1}|
        lt |dfrac{2x}{frac12}|
        =|4x|
        $        .
        To make
        $|4x| < epsilon$,
        you just need
        $|x| < dfrac{epsilon}{4}$.



        Therefore,
        if $|x| < min(dfrac14, dfrac{epsilon}{4})$,
        we have
        $|dfrac{2x}{2x+1}|<epsilon
        $        .



        If you want the best bounds,
        $|dfrac{2x}{2x+1}|<epsilon
        $
        is the same as
        $dfrac1{epsilon}
        lt |dfrac{2x+1}{2x}|
        = |1+dfrac{1}{2x}|
        $        .



        If $x > 0$,
        $|1+dfrac{1}{2x}|
        =1+dfrac{1}{2x}
        $
        so we want
        $dfrac1{epsilon}
        lt 1+dfrac{1}{2x}
        $
        or
        $dfrac1{epsilon}-1
        lt dfrac{1}{2x}
        $
        or,
        assuming $epsilon < 1$,
        $2x
        lt dfrac1{dfrac1{epsilon}-1}
        = dfrac{epsilon}{1-epsilon}
        $
        or
        $x
        ltdfrac{epsilon}{2(1-epsilon)}
        $        .



        If $x < 0$,
        to make
        $dfrac1{epsilon}
        lt |1+dfrac{1}{2x}|
        $
        be easy to work with,
        we need
        $1+dfrac{1}{2x} < 0$
        or
        $x > -dfrac12$.



        If this holds,
        then
        $|1+dfrac{1}{2x}|
        =-dfrac{1}{2x}-1
        $        ,
        so we want
        $-dfrac{1}{2x}-1
        gt dfrac1{epsilon}
        $
        or
        $-dfrac{1}{2x}
        gt dfrac1{epsilon}+1
        $
        or
        $-2x
        lt dfrac1{dfrac1{epsilon}+1}
        $
        or
        $-x
        lt dfrac{epsilon}{2(1+epsilon)}
        $        .



        I always prefer to get
        a simple bound
        which is not the best.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          So you want
          $|dfrac{2x}{2x+1}|<epsilon
          $          .



          There are two ways go here.



          For the first,
          note that you don't need
          the best bounds,
          just one that will work.



          Soppose
          $|x| < dfrac14$.
          Then
          $|2x| < dfrac12$
          so
          $|1+2x| > 1-2cdot dfrac14
          = dfrac12$          .
          Therefore
          $|dfrac{2x}{2x+1}|
          lt |dfrac{2x}{frac12}|
          =|4x|
          $          .
          To make
          $|4x| < epsilon$,
          you just need
          $|x| < dfrac{epsilon}{4}$.



          Therefore,
          if $|x| < min(dfrac14, dfrac{epsilon}{4})$,
          we have
          $|dfrac{2x}{2x+1}|<epsilon
          $          .



          If you want the best bounds,
          $|dfrac{2x}{2x+1}|<epsilon
          $
          is the same as
          $dfrac1{epsilon}
          lt |dfrac{2x+1}{2x}|
          = |1+dfrac{1}{2x}|
          $          .



          If $x > 0$,
          $|1+dfrac{1}{2x}|
          =1+dfrac{1}{2x}
          $
          so we want
          $dfrac1{epsilon}
          lt 1+dfrac{1}{2x}
          $
          or
          $dfrac1{epsilon}-1
          lt dfrac{1}{2x}
          $
          or,
          assuming $epsilon < 1$,
          $2x
          lt dfrac1{dfrac1{epsilon}-1}
          = dfrac{epsilon}{1-epsilon}
          $
          or
          $x
          ltdfrac{epsilon}{2(1-epsilon)}
          $          .



          If $x < 0$,
          to make
          $dfrac1{epsilon}
          lt |1+dfrac{1}{2x}|
          $
          be easy to work with,
          we need
          $1+dfrac{1}{2x} < 0$
          or
          $x > -dfrac12$.



          If this holds,
          then
          $|1+dfrac{1}{2x}|
          =-dfrac{1}{2x}-1
          $          ,
          so we want
          $-dfrac{1}{2x}-1
          gt dfrac1{epsilon}
          $
          or
          $-dfrac{1}{2x}
          gt dfrac1{epsilon}+1
          $
          or
          $-2x
          lt dfrac1{dfrac1{epsilon}+1}
          $
          or
          $-x
          lt dfrac{epsilon}{2(1+epsilon)}
          $          .



          I always prefer to get
          a simple bound
          which is not the best.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            So you want
            $|dfrac{2x}{2x+1}|<epsilon
            $            .



            There are two ways go here.



            For the first,
            note that you don't need
            the best bounds,
            just one that will work.



            Soppose
            $|x| < dfrac14$.
            Then
            $|2x| < dfrac12$
            so
            $|1+2x| > 1-2cdot dfrac14
            = dfrac12$            .
            Therefore
            $|dfrac{2x}{2x+1}|
            lt |dfrac{2x}{frac12}|
            =|4x|
            $            .
            To make
            $|4x| < epsilon$,
            you just need
            $|x| < dfrac{epsilon}{4}$.



            Therefore,
            if $|x| < min(dfrac14, dfrac{epsilon}{4})$,
            we have
            $|dfrac{2x}{2x+1}|<epsilon
            $            .



            If you want the best bounds,
            $|dfrac{2x}{2x+1}|<epsilon
            $
            is the same as
            $dfrac1{epsilon}
            lt |dfrac{2x+1}{2x}|
            = |1+dfrac{1}{2x}|
            $            .



            If $x > 0$,
            $|1+dfrac{1}{2x}|
            =1+dfrac{1}{2x}
            $
            so we want
            $dfrac1{epsilon}
            lt 1+dfrac{1}{2x}
            $
            or
            $dfrac1{epsilon}-1
            lt dfrac{1}{2x}
            $
            or,
            assuming $epsilon < 1$,
            $2x
            lt dfrac1{dfrac1{epsilon}-1}
            = dfrac{epsilon}{1-epsilon}
            $
            or
            $x
            ltdfrac{epsilon}{2(1-epsilon)}
            $            .



            If $x < 0$,
            to make
            $dfrac1{epsilon}
            lt |1+dfrac{1}{2x}|
            $
            be easy to work with,
            we need
            $1+dfrac{1}{2x} < 0$
            or
            $x > -dfrac12$.



            If this holds,
            then
            $|1+dfrac{1}{2x}|
            =-dfrac{1}{2x}-1
            $            ,
            so we want
            $-dfrac{1}{2x}-1
            gt dfrac1{epsilon}
            $
            or
            $-dfrac{1}{2x}
            gt dfrac1{epsilon}+1
            $
            or
            $-2x
            lt dfrac1{dfrac1{epsilon}+1}
            $
            or
            $-x
            lt dfrac{epsilon}{2(1+epsilon)}
            $            .



            I always prefer to get
            a simple bound
            which is not the best.






            share|cite|improve this answer









            $endgroup$



            So you want
            $|dfrac{2x}{2x+1}|<epsilon
            $            .



            There are two ways go here.



            For the first,
            note that you don't need
            the best bounds,
            just one that will work.



            Soppose
            $|x| < dfrac14$.
            Then
            $|2x| < dfrac12$
            so
            $|1+2x| > 1-2cdot dfrac14
            = dfrac12$            .
            Therefore
            $|dfrac{2x}{2x+1}|
            lt |dfrac{2x}{frac12}|
            =|4x|
            $            .
            To make
            $|4x| < epsilon$,
            you just need
            $|x| < dfrac{epsilon}{4}$.



            Therefore,
            if $|x| < min(dfrac14, dfrac{epsilon}{4})$,
            we have
            $|dfrac{2x}{2x+1}|<epsilon
            $            .



            If you want the best bounds,
            $|dfrac{2x}{2x+1}|<epsilon
            $
            is the same as
            $dfrac1{epsilon}
            lt |dfrac{2x+1}{2x}|
            = |1+dfrac{1}{2x}|
            $            .



            If $x > 0$,
            $|1+dfrac{1}{2x}|
            =1+dfrac{1}{2x}
            $
            so we want
            $dfrac1{epsilon}
            lt 1+dfrac{1}{2x}
            $
            or
            $dfrac1{epsilon}-1
            lt dfrac{1}{2x}
            $
            or,
            assuming $epsilon < 1$,
            $2x
            lt dfrac1{dfrac1{epsilon}-1}
            = dfrac{epsilon}{1-epsilon}
            $
            or
            $x
            ltdfrac{epsilon}{2(1-epsilon)}
            $            .



            If $x < 0$,
            to make
            $dfrac1{epsilon}
            lt |1+dfrac{1}{2x}|
            $
            be easy to work with,
            we need
            $1+dfrac{1}{2x} < 0$
            or
            $x > -dfrac12$.



            If this holds,
            then
            $|1+dfrac{1}{2x}|
            =-dfrac{1}{2x}-1
            $            ,
            so we want
            $-dfrac{1}{2x}-1
            gt dfrac1{epsilon}
            $
            or
            $-dfrac{1}{2x}
            gt dfrac1{epsilon}+1
            $
            or
            $-2x
            lt dfrac1{dfrac1{epsilon}+1}
            $
            or
            $-x
            lt dfrac{epsilon}{2(1+epsilon)}
            $            .



            I always prefer to get
            a simple bound
            which is not the best.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 19 '18 at 4:56









            marty cohenmarty cohen

            73.9k549128




            73.9k549128























                0












                $begingroup$

                $midfrac{-2x}{2x+1}mid=midfrac{-2}{2+frac1x}midltmidfrac2{frac 1x}midltepsilon$. So make $mid xmidltfrac{epsilon}2$. That is, $delta =frac {epsilon}2$.






                share|cite|improve this answer











                $endgroup$


















                  0












                  $begingroup$

                  $midfrac{-2x}{2x+1}mid=midfrac{-2}{2+frac1x}midltmidfrac2{frac 1x}midltepsilon$. So make $mid xmidltfrac{epsilon}2$. That is, $delta =frac {epsilon}2$.






                  share|cite|improve this answer











                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    $midfrac{-2x}{2x+1}mid=midfrac{-2}{2+frac1x}midltmidfrac2{frac 1x}midltepsilon$. So make $mid xmidltfrac{epsilon}2$. That is, $delta =frac {epsilon}2$.






                    share|cite|improve this answer











                    $endgroup$



                    $midfrac{-2x}{2x+1}mid=midfrac{-2}{2+frac1x}midltmidfrac2{frac 1x}midltepsilon$. So make $mid xmidltfrac{epsilon}2$. That is, $delta =frac {epsilon}2$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 19 '18 at 4:54

























                    answered Dec 19 '18 at 4:49









                    Chris CusterChris Custer

                    13.8k3827




                    13.8k3827























                        0












                        $begingroup$

                        For $quaddelta<dfrac{varepsilon}{2(1-varepsilon)}impliesquaddisplaystyleleft|dfrac{-2x}{2x+1}right|<dfrac{2delta}{2delta+1}=dfrac{1}{1+frac{1}{2delta}}<varepsilon$






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          For $quaddelta<dfrac{varepsilon}{2(1-varepsilon)}impliesquaddisplaystyleleft|dfrac{-2x}{2x+1}right|<dfrac{2delta}{2delta+1}=dfrac{1}{1+frac{1}{2delta}}<varepsilon$






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            For $quaddelta<dfrac{varepsilon}{2(1-varepsilon)}impliesquaddisplaystyleleft|dfrac{-2x}{2x+1}right|<dfrac{2delta}{2delta+1}=dfrac{1}{1+frac{1}{2delta}}<varepsilon$






                            share|cite|improve this answer











                            $endgroup$



                            For $quaddelta<dfrac{varepsilon}{2(1-varepsilon)}impliesquaddisplaystyleleft|dfrac{-2x}{2x+1}right|<dfrac{2delta}{2delta+1}=dfrac{1}{1+frac{1}{2delta}}<varepsilon$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 19 '18 at 5:10

























                            answered Dec 19 '18 at 4:48









                            Yadati KiranYadati Kiran

                            1,7911619




                            1,7911619






























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                                Wiesbaden

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                                Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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                                Marschland

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                                Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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                                Dieringhausen

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                                Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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