Prove the limit lim n→∞ $ {(2n^7 + 3n^5 + 4n) over (4n^7 − 7n^2 + 5)} $ = 1/2












1














Im trying to use the formal definition of a limit to prove
$lim limits_{x to ∞} $$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ = 1/2



I understand this problem is done backwards so I set up the equation



$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ - ${1over 2}$



$left|{(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)} - {1over 2} right|$ < ε



Both the denominator and numerator are always positive so we can drop the absolute value and simplify and rewrite



$ {(6n^3-7)over(8n^7-14n^2+10)}$ < $ {(6n^5-7n^2+8n+5)over(8n^7-14n^2+10)}$ < ε



$ {(1)over(4n^4)}$ = $ {(n^3)over(4n^7)}$ < $ {(3n^3)over(4n^7)}$ = $ {(6n^3)over(8n^7)}$ < $ {(6n^3)over(8n^7-14n^2+10)}$ < ε+7



$ {(1)over(ε+7)}$ < $4n^4$



$ {1over4ε+28^{1over4}}$ < n = N



Now I have my N so I can writ my proof:



Let ε, N > 0 and N < n. Suppose N = $ {1over4ε+28^{1over4}}$ . Then it must follow that:



$ {1over4ε+28^{1over4}}$ < n



$ {1over4ε+28}$ < $n^4$



$ {1overε+7}$ < $4n^4$



$ {1over 4n^4}$ < $ε+7$



from there on I dont know how to use my inequalities to get back to what I started with.






calculus limits analysis proof-writing







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    Im trying to use the formal definition of a limit to prove
    $lim limits_{x to ∞} $$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ = 1/2



    I understand this problem is done backwards so I set up the equation



    $ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ - ${1over 2}$



    $left|{(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)} - {1over 2} right|$ < ε



    Both the denominator and numerator are always positive so we can drop the absolute value and simplify and rewrite



    $ {(6n^3-7)over(8n^7-14n^2+10)}$ < $ {(6n^5-7n^2+8n+5)over(8n^7-14n^2+10)}$ < ε



    $ {(1)over(4n^4)}$ = $ {(n^3)over(4n^7)}$ < $ {(3n^3)over(4n^7)}$ = $ {(6n^3)over(8n^7)}$ < $ {(6n^3)over(8n^7-14n^2+10)}$ < ε+7



    $ {(1)over(ε+7)}$ < $4n^4$



    $ {1over4ε+28^{1over4}}$ < n = N



    Now I have my N so I can writ my proof:



    Let ε, N > 0 and N < n. Suppose N = $ {1over4ε+28^{1over4}}$ . Then it must follow that:



    $ {1over4ε+28^{1over4}}$ < n



    $ {1over4ε+28}$ < $n^4$



    $ {1overε+7}$ < $4n^4$



    $ {1over 4n^4}$ < $ε+7$



    from there on I dont know how to use my inequalities to get back to what I started with.






    calculus limits analysis proof-writing







    share|cite|improve this question

























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      1







      Im trying to use the formal definition of a limit to prove
      $lim limits_{x to ∞} $$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ = 1/2



      I understand this problem is done backwards so I set up the equation



      $ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ - ${1over 2}$



      $left|{(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)} - {1over 2} right|$ < ε



      Both the denominator and numerator are always positive so we can drop the absolute value and simplify and rewrite



      $ {(6n^3-7)over(8n^7-14n^2+10)}$ < $ {(6n^5-7n^2+8n+5)over(8n^7-14n^2+10)}$ < ε



      $ {(1)over(4n^4)}$ = $ {(n^3)over(4n^7)}$ < $ {(3n^3)over(4n^7)}$ = $ {(6n^3)over(8n^7)}$ < $ {(6n^3)over(8n^7-14n^2+10)}$ < ε+7



      $ {(1)over(ε+7)}$ < $4n^4$



      $ {1over4ε+28^{1over4}}$ < n = N



      Now I have my N so I can writ my proof:



      Let ε, N > 0 and N < n. Suppose N = $ {1over4ε+28^{1over4}}$ . Then it must follow that:



      $ {1over4ε+28^{1over4}}$ < n



      $ {1over4ε+28}$ < $n^4$



      $ {1overε+7}$ < $4n^4$



      $ {1over 4n^4}$ < $ε+7$



      from there on I dont know how to use my inequalities to get back to what I started with.






      calculus limits analysis proof-writing







      share|cite|improve this question













      Im trying to use the formal definition of a limit to prove
      $lim limits_{x to ∞} $$ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ = 1/2



      I understand this problem is done backwards so I set up the equation



      $ {(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)}$ - ${1over 2}$



      $left|{(2n^7 + 3n^5 + 4n)over(4n^7 − 7n^2 + 5)} - {1over 2} right|$ < ε



      Both the denominator and numerator are always positive so we can drop the absolute value and simplify and rewrite



      $ {(6n^3-7)over(8n^7-14n^2+10)}$ < $ {(6n^5-7n^2+8n+5)over(8n^7-14n^2+10)}$ < ε



      $ {(1)over(4n^4)}$ = $ {(n^3)over(4n^7)}$ < $ {(3n^3)over(4n^7)}$ = $ {(6n^3)over(8n^7)}$ < $ {(6n^3)over(8n^7-14n^2+10)}$ < ε+7



      $ {(1)over(ε+7)}$ < $4n^4$



      $ {1over4ε+28^{1over4}}$ < n = N



      Now I have my N so I can writ my proof:



      Let ε, N > 0 and N < n. Suppose N = $ {1over4ε+28^{1over4}}$ . Then it must follow that:



      $ {1over4ε+28^{1over4}}$ < n



      $ {1over4ε+28}$ < $n^4$



      $ {1overε+7}$ < $4n^4$



      $ {1over 4n^4}$ < $ε+7$



      from there on I dont know how to use my inequalities to get back to what I started with.






      calculus limits analysis proof-writing




      calculus limits analysis proof-writing






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      asked Nov 26 '18 at 19:24









      Malaikatu Kargbo

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          Note thatbegin{align}leftlvertfrac{2n^7+3n^5+4n}{4n^7-7n^2+5}-frac12rightrvert&=frac{lvert6n^5+7n^2+8n-5rvert}{lvert8n^7-14n^2+10rvert}\&leqslantfrac{26n^5}{8n^7-14n^2-10}\&=frac{26}{8n^2-14n^{-3}-10n^{-5}}\&leqslantfrac{26}{8n^2-24},end{align}if $n>1$. Now, use the fact thatbegin{align}frac{26}{8n^2-24}<varepsilon&iff8n^2-24>frac{26}varepsilon\&iff n>sqrt{frac{13}{4varepsilon}+3}.end{align}






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            I would rather use another approach:



            Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
            $$frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}=frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}$$
            Note that $$lim_{ntoinfty}Bigl(frac{1}{n^k}Bigr)=0;forall kinmathbb{N_{>0}}$$
            Thus $$lim_{ntoinfty}Biggl(frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}Biggr)=lim_{ntoinfty}Biggl(frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}Biggr)=frac{2+0+0}{4-0+0}=frac{1}{2}$$






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            • That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
              – KM101
              Nov 26 '18 at 19:49













            Your Answer





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            Note thatbegin{align}leftlvertfrac{2n^7+3n^5+4n}{4n^7-7n^2+5}-frac12rightrvert&=frac{lvert6n^5+7n^2+8n-5rvert}{lvert8n^7-14n^2+10rvert}\&leqslantfrac{26n^5}{8n^7-14n^2-10}\&=frac{26}{8n^2-14n^{-3}-10n^{-5}}\&leqslantfrac{26}{8n^2-24},end{align}if $n>1$. Now, use the fact thatbegin{align}frac{26}{8n^2-24}<varepsilon&iff8n^2-24>frac{26}varepsilon\&iff n>sqrt{frac{13}{4varepsilon}+3}.end{align}






            share|cite|improve this answer


























              1














              Note thatbegin{align}leftlvertfrac{2n^7+3n^5+4n}{4n^7-7n^2+5}-frac12rightrvert&=frac{lvert6n^5+7n^2+8n-5rvert}{lvert8n^7-14n^2+10rvert}\&leqslantfrac{26n^5}{8n^7-14n^2-10}\&=frac{26}{8n^2-14n^{-3}-10n^{-5}}\&leqslantfrac{26}{8n^2-24},end{align}if $n>1$. Now, use the fact thatbegin{align}frac{26}{8n^2-24}<varepsilon&iff8n^2-24>frac{26}varepsilon\&iff n>sqrt{frac{13}{4varepsilon}+3}.end{align}






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                Note thatbegin{align}leftlvertfrac{2n^7+3n^5+4n}{4n^7-7n^2+5}-frac12rightrvert&=frac{lvert6n^5+7n^2+8n-5rvert}{lvert8n^7-14n^2+10rvert}\&leqslantfrac{26n^5}{8n^7-14n^2-10}\&=frac{26}{8n^2-14n^{-3}-10n^{-5}}\&leqslantfrac{26}{8n^2-24},end{align}if $n>1$. Now, use the fact thatbegin{align}frac{26}{8n^2-24}<varepsilon&iff8n^2-24>frac{26}varepsilon\&iff n>sqrt{frac{13}{4varepsilon}+3}.end{align}






                share|cite|improve this answer












                Note thatbegin{align}leftlvertfrac{2n^7+3n^5+4n}{4n^7-7n^2+5}-frac12rightrvert&=frac{lvert6n^5+7n^2+8n-5rvert}{lvert8n^7-14n^2+10rvert}\&leqslantfrac{26n^5}{8n^7-14n^2-10}\&=frac{26}{8n^2-14n^{-3}-10n^{-5}}\&leqslantfrac{26}{8n^2-24},end{align}if $n>1$. Now, use the fact thatbegin{align}frac{26}{8n^2-24}<varepsilon&iff8n^2-24>frac{26}varepsilon\&iff n>sqrt{frac{13}{4varepsilon}+3}.end{align}







                share|cite|improve this answer












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                answered Nov 26 '18 at 19:43









                José Carlos Santos

                150k22121221




                150k22121221























                    0














                    I would rather use another approach:



                    Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
                    $$frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}=frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}$$
                    Note that $$lim_{ntoinfty}Bigl(frac{1}{n^k}Bigr)=0;forall kinmathbb{N_{>0}}$$
                    Thus $$lim_{ntoinfty}Biggl(frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}Biggr)=lim_{ntoinfty}Biggl(frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}Biggr)=frac{2+0+0}{4-0+0}=frac{1}{2}$$






                    share|cite|improve this answer





















                    • That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
                      – KM101
                      Nov 26 '18 at 19:49


















                    0














                    I would rather use another approach:



                    Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
                    $$frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}=frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}$$
                    Note that $$lim_{ntoinfty}Bigl(frac{1}{n^k}Bigr)=0;forall kinmathbb{N_{>0}}$$
                    Thus $$lim_{ntoinfty}Biggl(frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}Biggr)=lim_{ntoinfty}Biggl(frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}Biggr)=frac{2+0+0}{4-0+0}=frac{1}{2}$$






                    share|cite|improve this answer





















                    • That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
                      – KM101
                      Nov 26 '18 at 19:49
















                    0












                    0








                    0






                    I would rather use another approach:



                    Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
                    $$frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}=frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}$$
                    Note that $$lim_{ntoinfty}Bigl(frac{1}{n^k}Bigr)=0;forall kinmathbb{N_{>0}}$$
                    Thus $$lim_{ntoinfty}Biggl(frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}Biggr)=lim_{ntoinfty}Biggl(frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}Biggr)=frac{2+0+0}{4-0+0}=frac{1}{2}$$






                    share|cite|improve this answer












                    I would rather use another approach:



                    Divide first of all both, numerator and denominator by the biggest power $n^7$ as follows
                    $$frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}=frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}$$
                    Note that $$lim_{ntoinfty}Bigl(frac{1}{n^k}Bigr)=0;forall kinmathbb{N_{>0}}$$
                    Thus $$lim_{ntoinfty}Biggl(frac{(2n^7 + 3n^5 + 4n)}{(4n^7 − 7n^2 + 5)}Biggr)=lim_{ntoinfty}Biggl(frac{2+3frac{1}{n^2}+4frac{1}{n^6}}{4-7frac{1}{n^5}+5frac{1}{n^7}}Biggr)=frac{2+0+0}{4-0+0}=frac{1}{2}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 26 '18 at 19:41









                    Dr. Mathva

                    919316




                    919316












                    • That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
                      – KM101
                      Nov 26 '18 at 19:49




















                    • That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
                      – KM101
                      Nov 26 '18 at 19:49


















                    That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
                    – KM101
                    Nov 26 '18 at 19:49






                    That doesn’t really address the OP’s issue though, as the question asked is about the $epsilon-delta$ definition.
                    – KM101
                    Nov 26 '18 at 19:49




















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