Question regarding an inequality in Spivak's Calculus












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On page 92 in the 4th Edition of Spivak's Calculus, he mentions an inequality while discussing Limits.



|x-3| < 1, or 2 < x < 4



I understand how he got the upper (or < 4) portion of the inequality, but how was the number "2" determined. I might be missing a subtle point here regarding absolute values so I thought I'd ask. Appreciate it in advanced!










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  • $begingroup$
    $-1<x-3<1{{}}$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 15 at 17:27
















0












$begingroup$


On page 92 in the 4th Edition of Spivak's Calculus, he mentions an inequality while discussing Limits.



|x-3| < 1, or 2 < x < 4



I understand how he got the upper (or < 4) portion of the inequality, but how was the number "2" determined. I might be missing a subtle point here regarding absolute values so I thought I'd ask. Appreciate it in advanced!










share|cite|improve this question









$endgroup$












  • $begingroup$
    $-1<x-3<1{{}}$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 15 at 17:27














0












0








0





$begingroup$


On page 92 in the 4th Edition of Spivak's Calculus, he mentions an inequality while discussing Limits.



|x-3| < 1, or 2 < x < 4



I understand how he got the upper (or < 4) portion of the inequality, but how was the number "2" determined. I might be missing a subtle point here regarding absolute values so I thought I'd ask. Appreciate it in advanced!










share|cite|improve this question









$endgroup$




On page 92 in the 4th Edition of Spivak's Calculus, he mentions an inequality while discussing Limits.



|x-3| < 1, or 2 < x < 4



I understand how he got the upper (or < 4) portion of the inequality, but how was the number "2" determined. I might be missing a subtle point here regarding absolute values so I thought I'd ask. Appreciate it in advanced!







inequality






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asked Jan 15 at 17:25









Julian25Julian25

1104




1104












  • $begingroup$
    $-1<x-3<1{{}}$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 15 at 17:27


















  • $begingroup$
    $-1<x-3<1{{}}$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 15 at 17:27
















$begingroup$
$-1<x-3<1{{}}$.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:27




$begingroup$
$-1<x-3<1{{}}$.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:27










2 Answers
2






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oldest

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2












$begingroup$

$$|x-3| < 1$$ is equivalent to $$-1 < x-3 < 1.$$
It may be helpful to read "$|x-3|$" as "the distance between $x$ and $3$."






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah, that way of reading it is really helpful. Thanks!
    $endgroup$
    – Julian25
    Jan 15 at 17:28



















0












$begingroup$

By definition,
$$ |x-3| < 1 quadiffquad -1 < x-3 < 1. $$
Now add $3$ throughout.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

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    active

    oldest

    votes









    2












    $begingroup$

    $$|x-3| < 1$$ is equivalent to $$-1 < x-3 < 1.$$
    It may be helpful to read "$|x-3|$" as "the distance between $x$ and $3$."






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ah, that way of reading it is really helpful. Thanks!
      $endgroup$
      – Julian25
      Jan 15 at 17:28
















    2












    $begingroup$

    $$|x-3| < 1$$ is equivalent to $$-1 < x-3 < 1.$$
    It may be helpful to read "$|x-3|$" as "the distance between $x$ and $3$."






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ah, that way of reading it is really helpful. Thanks!
      $endgroup$
      – Julian25
      Jan 15 at 17:28














    2












    2








    2





    $begingroup$

    $$|x-3| < 1$$ is equivalent to $$-1 < x-3 < 1.$$
    It may be helpful to read "$|x-3|$" as "the distance between $x$ and $3$."






    share|cite|improve this answer









    $endgroup$



    $$|x-3| < 1$$ is equivalent to $$-1 < x-3 < 1.$$
    It may be helpful to read "$|x-3|$" as "the distance between $x$ and $3$."







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 15 at 17:26









    angryavianangryavian

    42.5k23481




    42.5k23481












    • $begingroup$
      Ah, that way of reading it is really helpful. Thanks!
      $endgroup$
      – Julian25
      Jan 15 at 17:28


















    • $begingroup$
      Ah, that way of reading it is really helpful. Thanks!
      $endgroup$
      – Julian25
      Jan 15 at 17:28
















    $begingroup$
    Ah, that way of reading it is really helpful. Thanks!
    $endgroup$
    – Julian25
    Jan 15 at 17:28




    $begingroup$
    Ah, that way of reading it is really helpful. Thanks!
    $endgroup$
    – Julian25
    Jan 15 at 17:28











    0












    $begingroup$

    By definition,
    $$ |x-3| < 1 quadiffquad -1 < x-3 < 1. $$
    Now add $3$ throughout.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      By definition,
      $$ |x-3| < 1 quadiffquad -1 < x-3 < 1. $$
      Now add $3$ throughout.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        By definition,
        $$ |x-3| < 1 quadiffquad -1 < x-3 < 1. $$
        Now add $3$ throughout.






        share|cite|improve this answer









        $endgroup$



        By definition,
        $$ |x-3| < 1 quadiffquad -1 < x-3 < 1. $$
        Now add $3$ throughout.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 17:26









        MisterRiemannMisterRiemann

        5,9151625




        5,9151625






























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