By using the definition of limit only, prove that $lim_{xrightarrow 0} frac1{3x+1} = 1$












1















By using the definition of a limit only, prove that



$lim_{xrightarrow 0} dfrac{1}{3x+1} = 1$




We need to find $$0<left|xright|<deltaquadimpliesquadleft|dfrac{1}{3x+1}-1right|<epsilon.$$
I have simplified $left|dfrac{1}{3x+1}-1right|$ down to $left|dfrac{-3x}{3x+1}right|$



Then since ${xrightarrow 0}$ we can assume $-1<x<1$ then $-2<3x+1<4$ which implies $$left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|<left|dfrac{-3delta}{4}right|=epsilon$$
No sure if the solution is correct






calculus limits epsilon-delta







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  • The step $$ left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
    – Ovi
    Aug 14 '17 at 21:59
















1















By using the definition of a limit only, prove that



$lim_{xrightarrow 0} dfrac{1}{3x+1} = 1$




We need to find $$0<left|xright|<deltaquadimpliesquadleft|dfrac{1}{3x+1}-1right|<epsilon.$$
I have simplified $left|dfrac{1}{3x+1}-1right|$ down to $left|dfrac{-3x}{3x+1}right|$



Then since ${xrightarrow 0}$ we can assume $-1<x<1$ then $-2<3x+1<4$ which implies $$left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|<left|dfrac{-3delta}{4}right|=epsilon$$
No sure if the solution is correct






calculus limits epsilon-delta







share|cite|improve this question
























  • The step $$ left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
    – Ovi
    Aug 14 '17 at 21:59














1












1








1








By using the definition of a limit only, prove that



$lim_{xrightarrow 0} dfrac{1}{3x+1} = 1$




We need to find $$0<left|xright|<deltaquadimpliesquadleft|dfrac{1}{3x+1}-1right|<epsilon.$$
I have simplified $left|dfrac{1}{3x+1}-1right|$ down to $left|dfrac{-3x}{3x+1}right|$



Then since ${xrightarrow 0}$ we can assume $-1<x<1$ then $-2<3x+1<4$ which implies $$left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|<left|dfrac{-3delta}{4}right|=epsilon$$
No sure if the solution is correct






calculus limits epsilon-delta







share|cite|improve this question
















By using the definition of a limit only, prove that



$lim_{xrightarrow 0} dfrac{1}{3x+1} = 1$




We need to find $$0<left|xright|<deltaquadimpliesquadleft|dfrac{1}{3x+1}-1right|<epsilon.$$
I have simplified $left|dfrac{1}{3x+1}-1right|$ down to $left|dfrac{-3x}{3x+1}right|$



Then since ${xrightarrow 0}$ we can assume $-1<x<1$ then $-2<3x+1<4$ which implies $$left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|<left|dfrac{-3delta}{4}right|=epsilon$$
No sure if the solution is correct






calculus limits epsilon-delta




calculus limits epsilon-delta






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edited Nov 27 at 12:15









Martin Sleziak

44.7k7115270




44.7k7115270










asked Aug 14 '17 at 21:50









Ben Jones

19111




19111












  • The step $$ left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
    – Ovi
    Aug 14 '17 at 21:59


















  • The step $$ left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
    – Ovi
    Aug 14 '17 at 21:59
















The step $$ left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
– Ovi
Aug 14 '17 at 21:59




The step $$ left|dfrac{-3x}{3x+1}right|<left|dfrac{-3x}{4}right|$$ is not correct. You said $3x+1<4$. If you are dividing $|-3x|$ by a bigger number, you get a smaller result.
– Ovi
Aug 14 '17 at 21:59










3 Answers
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2














If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $frac{1}{3x+1}$is unbounded.



Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < frac14$. Hence $delta < frac14$.



if $-frac14 < x < frac14$, $$-frac34+1 < 3x+1< frac34+1$$.



$$frac14 < 3x+1< frac74$$



$$left|frac{1}{3x+4} right| < 4$$



Hence $$12delta < epsilon$$






share|cite|improve this answer































    0














    Note that



    $$left|frac{1}{1+3x}-1right|=left|frac{3x}{1+3x}right| tag 1$$



    Now, we restrict $x$ such that $xin [-1/4,1/4]$. And with this restriction, it is easy to see that $1/4/ le 1+3x$. Using this in $(1)$ reveals that



    $$left|frac{1}{1+3x}-1right|le 12|x|tag 2$$



    Finally, given any $epsilon>0$,



    $$left|frac{1}{1+3x}-1right|<epsilon$$



    whenever $|x|<delta =minleft(frac14,frac{epsilon}{12}right)$.






    share|cite|improve this answer





















    • Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
      – Mark Viola
      Aug 18 '17 at 16:00



















    0














    No. If we assume $-1 < x < 1$, then $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $left|dfrac {1}{3x+1}right|$ will be unbounded.



    If you want to put an upper bound on $left|dfrac {1}{3x+1}right|$, then you are going to need something like $3x + 1 > frac 13$, which is equivalent to
    $x > -frac 29$.



    So, $|x| < frac 29 implies 3x+1 > frac 13 implies
    left|dfrac {1}{3x+1}right|<3$.



    So, then, $left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|
    < |9x|$



    To make that less than $epsilon$, we need to have $|x|<frac 19 epsilon$.



    So we need to have $delta < frac 19epsilon$ and $delta < frac 29$.



    Hence we define $delta = min(frac 19epsilon, frac 29)$






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      2














      If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $frac{1}{3x+1}$is unbounded.



      Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < frac14$. Hence $delta < frac14$.



      if $-frac14 < x < frac14$, $$-frac34+1 < 3x+1< frac34+1$$.



      $$frac14 < 3x+1< frac74$$



      $$left|frac{1}{3x+4} right| < 4$$



      Hence $$12delta < epsilon$$






      share|cite|improve this answer




























        2














        If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $frac{1}{3x+1}$is unbounded.



        Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < frac14$. Hence $delta < frac14$.



        if $-frac14 < x < frac14$, $$-frac34+1 < 3x+1< frac34+1$$.



        $$frac14 < 3x+1< frac74$$



        $$left|frac{1}{3x+4} right| < 4$$



        Hence $$12delta < epsilon$$






        share|cite|improve this answer


























          2












          2








          2






          If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $frac{1}{3x+1}$is unbounded.



          Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < frac14$. Hence $delta < frac14$.



          if $-frac14 < x < frac14$, $$-frac34+1 < 3x+1< frac34+1$$.



          $$frac14 < 3x+1< frac74$$



          $$left|frac{1}{3x+4} right| < 4$$



          Hence $$12delta < epsilon$$






          share|cite|improve this answer














          If we focus on $-1<x<1$ and end up with $-2<3x+1 < 4$, $frac{1}{3x+1}$is unbounded.



          Rather than focusing on $-1 < x < 1$, focus on a smaller interval, for example $|x| < frac14$. Hence $delta < frac14$.



          if $-frac14 < x < frac14$, $$-frac34+1 < 3x+1< frac34+1$$.



          $$frac14 < 3x+1< frac74$$



          $$left|frac{1}{3x+4} right| < 4$$



          Hence $$12delta < epsilon$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 14 '17 at 22:06

























          answered Aug 14 '17 at 21:59









          Siong Thye Goh

          99.1k1464117




          99.1k1464117























              0














              Note that



              $$left|frac{1}{1+3x}-1right|=left|frac{3x}{1+3x}right| tag 1$$



              Now, we restrict $x$ such that $xin [-1/4,1/4]$. And with this restriction, it is easy to see that $1/4/ le 1+3x$. Using this in $(1)$ reveals that



              $$left|frac{1}{1+3x}-1right|le 12|x|tag 2$$



              Finally, given any $epsilon>0$,



              $$left|frac{1}{1+3x}-1right|<epsilon$$



              whenever $|x|<delta =minleft(frac14,frac{epsilon}{12}right)$.






              share|cite|improve this answer





















              • Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
                – Mark Viola
                Aug 18 '17 at 16:00
















              0














              Note that



              $$left|frac{1}{1+3x}-1right|=left|frac{3x}{1+3x}right| tag 1$$



              Now, we restrict $x$ such that $xin [-1/4,1/4]$. And with this restriction, it is easy to see that $1/4/ le 1+3x$. Using this in $(1)$ reveals that



              $$left|frac{1}{1+3x}-1right|le 12|x|tag 2$$



              Finally, given any $epsilon>0$,



              $$left|frac{1}{1+3x}-1right|<epsilon$$



              whenever $|x|<delta =minleft(frac14,frac{epsilon}{12}right)$.






              share|cite|improve this answer





















              • Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
                – Mark Viola
                Aug 18 '17 at 16:00














              0












              0








              0






              Note that



              $$left|frac{1}{1+3x}-1right|=left|frac{3x}{1+3x}right| tag 1$$



              Now, we restrict $x$ such that $xin [-1/4,1/4]$. And with this restriction, it is easy to see that $1/4/ le 1+3x$. Using this in $(1)$ reveals that



              $$left|frac{1}{1+3x}-1right|le 12|x|tag 2$$



              Finally, given any $epsilon>0$,



              $$left|frac{1}{1+3x}-1right|<epsilon$$



              whenever $|x|<delta =minleft(frac14,frac{epsilon}{12}right)$.






              share|cite|improve this answer












              Note that



              $$left|frac{1}{1+3x}-1right|=left|frac{3x}{1+3x}right| tag 1$$



              Now, we restrict $x$ such that $xin [-1/4,1/4]$. And with this restriction, it is easy to see that $1/4/ le 1+3x$. Using this in $(1)$ reveals that



              $$left|frac{1}{1+3x}-1right|le 12|x|tag 2$$



              Finally, given any $epsilon>0$,



              $$left|frac{1}{1+3x}-1right|<epsilon$$



              whenever $|x|<delta =minleft(frac14,frac{epsilon}{12}right)$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Aug 14 '17 at 21:59









              Mark Viola

              130k1274170




              130k1274170












              • Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
                – Mark Viola
                Aug 18 '17 at 16:00


















              • Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
                – Mark Viola
                Aug 18 '17 at 16:00
















              Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
              – Mark Viola
              Aug 18 '17 at 16:00




              Please let me know how I can improve my answer. I really want to give you the best answer I can. And once you've accrues enough reputation points, you can up vote answers as you see fit.
              – Mark Viola
              Aug 18 '17 at 16:00











              0














              No. If we assume $-1 < x < 1$, then $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $left|dfrac {1}{3x+1}right|$ will be unbounded.



              If you want to put an upper bound on $left|dfrac {1}{3x+1}right|$, then you are going to need something like $3x + 1 > frac 13$, which is equivalent to
              $x > -frac 29$.



              So, $|x| < frac 29 implies 3x+1 > frac 13 implies
              left|dfrac {1}{3x+1}right|<3$.



              So, then, $left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|
              < |9x|$



              To make that less than $epsilon$, we need to have $|x|<frac 19 epsilon$.



              So we need to have $delta < frac 19epsilon$ and $delta < frac 29$.



              Hence we define $delta = min(frac 19epsilon, frac 29)$






              share|cite|improve this answer


























                0














                No. If we assume $-1 < x < 1$, then $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $left|dfrac {1}{3x+1}right|$ will be unbounded.



                If you want to put an upper bound on $left|dfrac {1}{3x+1}right|$, then you are going to need something like $3x + 1 > frac 13$, which is equivalent to
                $x > -frac 29$.



                So, $|x| < frac 29 implies 3x+1 > frac 13 implies
                left|dfrac {1}{3x+1}right|<3$.



                So, then, $left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|
                < |9x|$



                To make that less than $epsilon$, we need to have $|x|<frac 19 epsilon$.



                So we need to have $delta < frac 19epsilon$ and $delta < frac 29$.



                Hence we define $delta = min(frac 19epsilon, frac 29)$






                share|cite|improve this answer
























                  0












                  0








                  0






                  No. If we assume $-1 < x < 1$, then $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $left|dfrac {1}{3x+1}right|$ will be unbounded.



                  If you want to put an upper bound on $left|dfrac {1}{3x+1}right|$, then you are going to need something like $3x + 1 > frac 13$, which is equivalent to
                  $x > -frac 29$.



                  So, $|x| < frac 29 implies 3x+1 > frac 13 implies
                  left|dfrac {1}{3x+1}right|<3$.



                  So, then, $left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|
                  < |9x|$



                  To make that less than $epsilon$, we need to have $|x|<frac 19 epsilon$.



                  So we need to have $delta < frac 19epsilon$ and $delta < frac 29$.



                  Hence we define $delta = min(frac 19epsilon, frac 29)$






                  share|cite|improve this answer












                  No. If we assume $-1 < x < 1$, then $-2 < 3x + 1 < 4$. Since $3x+1=0$ is within those limits, then $left|dfrac {1}{3x+1}right|$ will be unbounded.



                  If you want to put an upper bound on $left|dfrac {1}{3x+1}right|$, then you are going to need something like $3x + 1 > frac 13$, which is equivalent to
                  $x > -frac 29$.



                  So, $|x| < frac 29 implies 3x+1 > frac 13 implies
                  left|dfrac {1}{3x+1}right|<3$.



                  So, then, $left|dfrac{1}{3x+1}-1right|=left|dfrac{-3x}{3x+1}right|
                  < |9x|$



                  To make that less than $epsilon$, we need to have $|x|<frac 19 epsilon$.



                  So we need to have $delta < frac 19epsilon$ and $delta < frac 29$.



                  Hence we define $delta = min(frac 19epsilon, frac 29)$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 14 '17 at 23:21









                  steven gregory

                  17.7k32257




                  17.7k32257






























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