Limit $limlimits_{xrightarrow 5}x^2=25$











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From my book:




6. With regard to the limit $displaystyle lim_{xrightarrow 5}x^2=25$,



(a) Show that $left |x^2-25 right |<11left |x-5 right |$ if $4<x<6$



(b) Find a $delta$ such that $left |x^2-25 right |<10^{-3}$ if $0<left |x-5 right |<delta$



(c) Give a rigorous proof of the limit by showing that $left |x^2-25 right |<epsilon$ if $0<left|x-5right|<delta$, where $delta$ is the smaller of $frac{epsilon}{11}$ and $1$.





For (a):



$|x^2-25|<11|x-5|$



$|x+5||x-5|<11|x-5|$



$|x+5|<11$



$-16<x<6$



And ignoring the interval $[-16,4)$, $4<x<6$




For (b) and (c) I am unsure.






calculus limits epsilon-delta







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  • 1




    Your (a) is fine. So now use it. Suppose $|x-5|<delta$. What can you conclude about $|x^2-25|$?
    – almagest
    Sep 8 '14 at 19:24










  • Just a small nitpick: You have to consider the case $x=5$ separately, because otherwise you are dividing both sides by zero when you go from $|x+5||x-5|<11|x-5|$ to $|x+5|<11$.
    – TonyK
    Sep 8 '14 at 19:34










  • I think your (a) is a bit of an overkill: it enough to take the upper bound on $x: |x+5|<|6+5|=11$
    – Alex
    Sep 8 '14 at 21:27















up vote
2
down vote

favorite












From my book:




6. With regard to the limit $displaystyle lim_{xrightarrow 5}x^2=25$,



(a) Show that $left |x^2-25 right |<11left |x-5 right |$ if $4<x<6$



(b) Find a $delta$ such that $left |x^2-25 right |<10^{-3}$ if $0<left |x-5 right |<delta$



(c) Give a rigorous proof of the limit by showing that $left |x^2-25 right |<epsilon$ if $0<left|x-5right|<delta$, where $delta$ is the smaller of $frac{epsilon}{11}$ and $1$.





For (a):



$|x^2-25|<11|x-5|$



$|x+5||x-5|<11|x-5|$



$|x+5|<11$



$-16<x<6$



And ignoring the interval $[-16,4)$, $4<x<6$




For (b) and (c) I am unsure.






calculus limits epsilon-delta







share|cite|improve this question




















  • 1




    Your (a) is fine. So now use it. Suppose $|x-5|<delta$. What can you conclude about $|x^2-25|$?
    – almagest
    Sep 8 '14 at 19:24










  • Just a small nitpick: You have to consider the case $x=5$ separately, because otherwise you are dividing both sides by zero when you go from $|x+5||x-5|<11|x-5|$ to $|x+5|<11$.
    – TonyK
    Sep 8 '14 at 19:34










  • I think your (a) is a bit of an overkill: it enough to take the upper bound on $x: |x+5|<|6+5|=11$
    – Alex
    Sep 8 '14 at 21:27













up vote
2
down vote

favorite









up vote
2
down vote

favorite











From my book:




6. With regard to the limit $displaystyle lim_{xrightarrow 5}x^2=25$,



(a) Show that $left |x^2-25 right |<11left |x-5 right |$ if $4<x<6$



(b) Find a $delta$ such that $left |x^2-25 right |<10^{-3}$ if $0<left |x-5 right |<delta$



(c) Give a rigorous proof of the limit by showing that $left |x^2-25 right |<epsilon$ if $0<left|x-5right|<delta$, where $delta$ is the smaller of $frac{epsilon}{11}$ and $1$.





For (a):



$|x^2-25|<11|x-5|$



$|x+5||x-5|<11|x-5|$



$|x+5|<11$



$-16<x<6$



And ignoring the interval $[-16,4)$, $4<x<6$




For (b) and (c) I am unsure.






calculus limits epsilon-delta







share|cite|improve this question















From my book:




6. With regard to the limit $displaystyle lim_{xrightarrow 5}x^2=25$,



(a) Show that $left |x^2-25 right |<11left |x-5 right |$ if $4<x<6$



(b) Find a $delta$ such that $left |x^2-25 right |<10^{-3}$ if $0<left |x-5 right |<delta$



(c) Give a rigorous proof of the limit by showing that $left |x^2-25 right |<epsilon$ if $0<left|x-5right|<delta$, where $delta$ is the smaller of $frac{epsilon}{11}$ and $1$.





For (a):



$|x^2-25|<11|x-5|$



$|x+5||x-5|<11|x-5|$



$|x+5|<11$



$-16<x<6$



And ignoring the interval $[-16,4)$, $4<x<6$




For (b) and (c) I am unsure.






calculus limits epsilon-delta




calculus limits epsilon-delta






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edited Nov 27 at 12:14









Martin Sleziak

44.6k7115269




44.6k7115269










asked Sep 8 '14 at 19:21









Gᴇᴏᴍᴇᴛᴇʀ

4752621




4752621








  • 1




    Your (a) is fine. So now use it. Suppose $|x-5|<delta$. What can you conclude about $|x^2-25|$?
    – almagest
    Sep 8 '14 at 19:24










  • Just a small nitpick: You have to consider the case $x=5$ separately, because otherwise you are dividing both sides by zero when you go from $|x+5||x-5|<11|x-5|$ to $|x+5|<11$.
    – TonyK
    Sep 8 '14 at 19:34










  • I think your (a) is a bit of an overkill: it enough to take the upper bound on $x: |x+5|<|6+5|=11$
    – Alex
    Sep 8 '14 at 21:27














  • 1




    Your (a) is fine. So now use it. Suppose $|x-5|<delta$. What can you conclude about $|x^2-25|$?
    – almagest
    Sep 8 '14 at 19:24










  • Just a small nitpick: You have to consider the case $x=5$ separately, because otherwise you are dividing both sides by zero when you go from $|x+5||x-5|<11|x-5|$ to $|x+5|<11$.
    – TonyK
    Sep 8 '14 at 19:34










  • I think your (a) is a bit of an overkill: it enough to take the upper bound on $x: |x+5|<|6+5|=11$
    – Alex
    Sep 8 '14 at 21:27








1




1




Your (a) is fine. So now use it. Suppose $|x-5|<delta$. What can you conclude about $|x^2-25|$?
– almagest
Sep 8 '14 at 19:24




Your (a) is fine. So now use it. Suppose $|x-5|<delta$. What can you conclude about $|x^2-25|$?
– almagest
Sep 8 '14 at 19:24












Just a small nitpick: You have to consider the case $x=5$ separately, because otherwise you are dividing both sides by zero when you go from $|x+5||x-5|<11|x-5|$ to $|x+5|<11$.
– TonyK
Sep 8 '14 at 19:34




Just a small nitpick: You have to consider the case $x=5$ separately, because otherwise you are dividing both sides by zero when you go from $|x+5||x-5|<11|x-5|$ to $|x+5|<11$.
– TonyK
Sep 8 '14 at 19:34












I think your (a) is a bit of an overkill: it enough to take the upper bound on $x: |x+5|<|6+5|=11$
– Alex
Sep 8 '14 at 21:27




I think your (a) is a bit of an overkill: it enough to take the upper bound on $x: |x+5|<|6+5|=11$
– Alex
Sep 8 '14 at 21:27










2 Answers
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Hint: let's find a $delta'$ so that $11|x-5| < 10^{-3}$ whenever $|x-5| < delta'$. If we select $delta = min{delta',1}$, then we can guarantee that if $|x-5| < delta$, then (since $x$ is necessarily between $4$ and $6$),
$$
|x^2 - 5| < 11|x - 5| < 10^{-3}
$$
For (c), all you should need to do is substitute $epsilon$ for $10^{-3}$.






share|cite|improve this answer




























    up vote
    0
    down vote













    If $0<|x-5|<delta$ we have $$5-delta<x<5+deltato 25-10delta+delta^2<x^2<25+10delta+delta^2$$therefore $$-11delta<-10delta<x^2-25<11delta\ |x^2-25|<11delta$$ for any $0<delta<1$. Now take $deltale min{1,{10^{-3}over 11}}={10^{-3}over 11}$ (for example $delta=0.0005$) and you are done.



    For part c, just easily substitute $10^{-3}$ with $epsilon $ and complete the proof.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      up vote
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      Hint: let's find a $delta'$ so that $11|x-5| < 10^{-3}$ whenever $|x-5| < delta'$. If we select $delta = min{delta',1}$, then we can guarantee that if $|x-5| < delta$, then (since $x$ is necessarily between $4$ and $6$),
      $$
      |x^2 - 5| < 11|x - 5| < 10^{-3}
      $$
      For (c), all you should need to do is substitute $epsilon$ for $10^{-3}$.






      share|cite|improve this answer

























        up vote
        0
        down vote













        Hint: let's find a $delta'$ so that $11|x-5| < 10^{-3}$ whenever $|x-5| < delta'$. If we select $delta = min{delta',1}$, then we can guarantee that if $|x-5| < delta$, then (since $x$ is necessarily between $4$ and $6$),
        $$
        |x^2 - 5| < 11|x - 5| < 10^{-3}
        $$
        For (c), all you should need to do is substitute $epsilon$ for $10^{-3}$.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          Hint: let's find a $delta'$ so that $11|x-5| < 10^{-3}$ whenever $|x-5| < delta'$. If we select $delta = min{delta',1}$, then we can guarantee that if $|x-5| < delta$, then (since $x$ is necessarily between $4$ and $6$),
          $$
          |x^2 - 5| < 11|x - 5| < 10^{-3}
          $$
          For (c), all you should need to do is substitute $epsilon$ for $10^{-3}$.






          share|cite|improve this answer












          Hint: let's find a $delta'$ so that $11|x-5| < 10^{-3}$ whenever $|x-5| < delta'$. If we select $delta = min{delta',1}$, then we can guarantee that if $|x-5| < delta$, then (since $x$ is necessarily between $4$ and $6$),
          $$
          |x^2 - 5| < 11|x - 5| < 10^{-3}
          $$
          For (c), all you should need to do is substitute $epsilon$ for $10^{-3}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 8 '14 at 19:26









          Omnomnomnom

          125k788176




          125k788176






















              up vote
              0
              down vote













              If $0<|x-5|<delta$ we have $$5-delta<x<5+deltato 25-10delta+delta^2<x^2<25+10delta+delta^2$$therefore $$-11delta<-10delta<x^2-25<11delta\ |x^2-25|<11delta$$ for any $0<delta<1$. Now take $deltale min{1,{10^{-3}over 11}}={10^{-3}over 11}$ (for example $delta=0.0005$) and you are done.



              For part c, just easily substitute $10^{-3}$ with $epsilon $ and complete the proof.






              share|cite|improve this answer

























                up vote
                0
                down vote













                If $0<|x-5|<delta$ we have $$5-delta<x<5+deltato 25-10delta+delta^2<x^2<25+10delta+delta^2$$therefore $$-11delta<-10delta<x^2-25<11delta\ |x^2-25|<11delta$$ for any $0<delta<1$. Now take $deltale min{1,{10^{-3}over 11}}={10^{-3}over 11}$ (for example $delta=0.0005$) and you are done.



                For part c, just easily substitute $10^{-3}$ with $epsilon $ and complete the proof.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  If $0<|x-5|<delta$ we have $$5-delta<x<5+deltato 25-10delta+delta^2<x^2<25+10delta+delta^2$$therefore $$-11delta<-10delta<x^2-25<11delta\ |x^2-25|<11delta$$ for any $0<delta<1$. Now take $deltale min{1,{10^{-3}over 11}}={10^{-3}over 11}$ (for example $delta=0.0005$) and you are done.



                  For part c, just easily substitute $10^{-3}$ with $epsilon $ and complete the proof.






                  share|cite|improve this answer












                  If $0<|x-5|<delta$ we have $$5-delta<x<5+deltato 25-10delta+delta^2<x^2<25+10delta+delta^2$$therefore $$-11delta<-10delta<x^2-25<11delta\ |x^2-25|<11delta$$ for any $0<delta<1$. Now take $deltale min{1,{10^{-3}over 11}}={10^{-3}over 11}$ (for example $delta=0.0005$) and you are done.



                  For part c, just easily substitute $10^{-3}$ with $epsilon $ and complete the proof.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 at 12:27









                  Mostafa Ayaz

                  13.5k3836




                  13.5k3836






























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