Evaluating the limit $lim_{x to infty}left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)$












-3














I wanna know how to do this limit



$lim_{x to infty}left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)$






calculus limits infinity







share|cite|improve this question
























  • Try multiplying and dividing by the conjugate root to eliminate the roots.
    – D.B.
    Nov 22 at 4:37










  • thats what i did, but i got stuck
    – Franco Cabrera
    Nov 22 at 4:41










  • math.stackexchange.com/questions/2959619/…
    – lab bhattacharjee
    Nov 22 at 4:41










  • Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
    – Ovi
    Nov 22 at 4:49
















-3














I wanna know how to do this limit



$lim_{x to infty}left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)$






calculus limits infinity







share|cite|improve this question
























  • Try multiplying and dividing by the conjugate root to eliminate the roots.
    – D.B.
    Nov 22 at 4:37










  • thats what i did, but i got stuck
    – Franco Cabrera
    Nov 22 at 4:41










  • math.stackexchange.com/questions/2959619/…
    – lab bhattacharjee
    Nov 22 at 4:41










  • Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
    – Ovi
    Nov 22 at 4:49














-3












-3








-3







I wanna know how to do this limit



$lim_{x to infty}left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)$






calculus limits infinity







share|cite|improve this question















I wanna know how to do this limit



$lim_{x to infty}left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)$






calculus limits infinity




calculus limits infinity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 4:37

























asked Nov 22 at 4:32









Franco Cabrera

64




64












  • Try multiplying and dividing by the conjugate root to eliminate the roots.
    – D.B.
    Nov 22 at 4:37










  • thats what i did, but i got stuck
    – Franco Cabrera
    Nov 22 at 4:41










  • math.stackexchange.com/questions/2959619/…
    – lab bhattacharjee
    Nov 22 at 4:41










  • Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
    – Ovi
    Nov 22 at 4:49


















  • Try multiplying and dividing by the conjugate root to eliminate the roots.
    – D.B.
    Nov 22 at 4:37










  • thats what i did, but i got stuck
    – Franco Cabrera
    Nov 22 at 4:41










  • math.stackexchange.com/questions/2959619/…
    – lab bhattacharjee
    Nov 22 at 4:41










  • Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
    – Ovi
    Nov 22 at 4:49
















Try multiplying and dividing by the conjugate root to eliminate the roots.
– D.B.
Nov 22 at 4:37




Try multiplying and dividing by the conjugate root to eliminate the roots.
– D.B.
Nov 22 at 4:37












thats what i did, but i got stuck
– Franco Cabrera
Nov 22 at 4:41




thats what i did, but i got stuck
– Franco Cabrera
Nov 22 at 4:41












math.stackexchange.com/questions/2959619/…
– lab bhattacharjee
Nov 22 at 4:41




math.stackexchange.com/questions/2959619/…
– lab bhattacharjee
Nov 22 at 4:41












Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
– Ovi
Nov 22 at 4:49




Remember that intuitively, when you work with limits of polynomials, you can often just replace the polynomial with it's leading term and get the right answer. In this case you get $0$
– Ovi
Nov 22 at 4:49










3 Answers
3






active

oldest

votes


















1














First of all, rationalise the expression by multiplying numerator and denominator by$displaystyle sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}$
begin{equation*}
end{equation*}
We get
begin{gather*}
lim xrightarrow infty frac{-x^{3} -15x^{2} +6}{sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}}\
end{gather*}
Divide numerator and denominator by $displaystyle x^{2}$
begin{equation*}
lim xrightarrow infty frac{-x-15+frac{6}{x^{2}}}{sqrt{1-frac{1}{x} +frac{1}{x^{4}}} +sqrt{1+frac{15}{x^{2}} -frac{5}{x^{4}}}}
end{equation*}



Clearly the given expression tends to$displaystyle -infty $ as $displaystyle xrightarrow infty $
begin{equation*}
end{equation*}






share|cite|improve this answer





























    0














    You should multiply by the conjugate of the expression, then proceed.



    begin{align}
    frac{left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}\
    end{align}
    begin{align}
    frac{(x^4 -x^3+1)-(x^4+15x^2-5)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}
    end{align}



    Operate and calculate the limit.






    share|cite|improve this answer























    • negative! that's cool!
      – Christopher Marley
      Nov 22 at 4:52










    • What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
      – Alberto Torrejon Valenzuela
      Nov 22 at 4:54










    • and then, should i evaluate de sign of de function to see the lateral limits?
      – Franco Cabrera
      Nov 22 at 4:54










    • No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
      – Alberto Torrejon Valenzuela
      Nov 22 at 4:56










    • Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
      – Franco Cabrera
      Nov 22 at 4:56



















    -1














    Another manipulation:



    Let $a,b >0$, real.



    $a-b=(√a-√b)(√a+√b)$, then



    $√a-√b = dfrac{a-b}{√a+√b}.$



    $a: = (x^4-x^3+1)^{1/2},$
    $b=(x^4+15x^2-5)^{1/2}.$



    $small{f(x):=}$



    $small {=dfrac{-x^3-15x^2+6}{(x^4-x^3+1)^{1/2}+(x^4+15x^2-5)^{1/2}}}$



    Note : For large x (positive) the numerator is negative, the denominator positive: Increasing the denominator makes the fraction bigger.



    Hence



    $f(x) lt $



    $dfrac{-x^3+6}{(x^4+1)^{1/2}+(x^4+15x^2)^{1/2}} lt $



    $dfrac{-x^3+6}{√2x^2+√2x^2}=$



    $dfrac{-x^3+6}{(2√2)x^2}.$



    Take the limit.






    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









      1














      First of all, rationalise the expression by multiplying numerator and denominator by$displaystyle sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}$
      begin{equation*}
      end{equation*}
      We get
      begin{gather*}
      lim xrightarrow infty frac{-x^{3} -15x^{2} +6}{sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}}\
      end{gather*}
      Divide numerator and denominator by $displaystyle x^{2}$
      begin{equation*}
      lim xrightarrow infty frac{-x-15+frac{6}{x^{2}}}{sqrt{1-frac{1}{x} +frac{1}{x^{4}}} +sqrt{1+frac{15}{x^{2}} -frac{5}{x^{4}}}}
      end{equation*}



      Clearly the given expression tends to$displaystyle -infty $ as $displaystyle xrightarrow infty $
      begin{equation*}
      end{equation*}






      share|cite|improve this answer


























        1














        First of all, rationalise the expression by multiplying numerator and denominator by$displaystyle sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}$
        begin{equation*}
        end{equation*}
        We get
        begin{gather*}
        lim xrightarrow infty frac{-x^{3} -15x^{2} +6}{sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}}\
        end{gather*}
        Divide numerator and denominator by $displaystyle x^{2}$
        begin{equation*}
        lim xrightarrow infty frac{-x-15+frac{6}{x^{2}}}{sqrt{1-frac{1}{x} +frac{1}{x^{4}}} +sqrt{1+frac{15}{x^{2}} -frac{5}{x^{4}}}}
        end{equation*}



        Clearly the given expression tends to$displaystyle -infty $ as $displaystyle xrightarrow infty $
        begin{equation*}
        end{equation*}






        share|cite|improve this answer
























          1












          1








          1






          First of all, rationalise the expression by multiplying numerator and denominator by$displaystyle sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}$
          begin{equation*}
          end{equation*}
          We get
          begin{gather*}
          lim xrightarrow infty frac{-x^{3} -15x^{2} +6}{sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}}\
          end{gather*}
          Divide numerator and denominator by $displaystyle x^{2}$
          begin{equation*}
          lim xrightarrow infty frac{-x-15+frac{6}{x^{2}}}{sqrt{1-frac{1}{x} +frac{1}{x^{4}}} +sqrt{1+frac{15}{x^{2}} -frac{5}{x^{4}}}}
          end{equation*}



          Clearly the given expression tends to$displaystyle -infty $ as $displaystyle xrightarrow infty $
          begin{equation*}
          end{equation*}






          share|cite|improve this answer












          First of all, rationalise the expression by multiplying numerator and denominator by$displaystyle sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}$
          begin{equation*}
          end{equation*}
          We get
          begin{gather*}
          lim xrightarrow infty frac{-x^{3} -15x^{2} +6}{sqrt{x^{4} -x^{3} +1} + sqrt{x^{4} +15x^{2} -5}}\
          end{gather*}
          Divide numerator and denominator by $displaystyle x^{2}$
          begin{equation*}
          lim xrightarrow infty frac{-x-15+frac{6}{x^{2}}}{sqrt{1-frac{1}{x} +frac{1}{x^{4}}} +sqrt{1+frac{15}{x^{2}} -frac{5}{x^{4}}}}
          end{equation*}



          Clearly the given expression tends to$displaystyle -infty $ as $displaystyle xrightarrow infty $
          begin{equation*}
          end{equation*}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 5:49









          Dikshit Gautam

          795




          795























              0














              You should multiply by the conjugate of the expression, then proceed.



              begin{align}
              frac{left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}\
              end{align}
              begin{align}
              frac{(x^4 -x^3+1)-(x^4+15x^2-5)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}
              end{align}



              Operate and calculate the limit.






              share|cite|improve this answer























              • negative! that's cool!
                – Christopher Marley
                Nov 22 at 4:52










              • What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
                – Alberto Torrejon Valenzuela
                Nov 22 at 4:54










              • and then, should i evaluate de sign of de function to see the lateral limits?
                – Franco Cabrera
                Nov 22 at 4:54










              • No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
                – Alberto Torrejon Valenzuela
                Nov 22 at 4:56










              • Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
                – Franco Cabrera
                Nov 22 at 4:56
















              0














              You should multiply by the conjugate of the expression, then proceed.



              begin{align}
              frac{left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}\
              end{align}
              begin{align}
              frac{(x^4 -x^3+1)-(x^4+15x^2-5)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}
              end{align}



              Operate and calculate the limit.






              share|cite|improve this answer























              • negative! that's cool!
                – Christopher Marley
                Nov 22 at 4:52










              • What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
                – Alberto Torrejon Valenzuela
                Nov 22 at 4:54










              • and then, should i evaluate de sign of de function to see the lateral limits?
                – Franco Cabrera
                Nov 22 at 4:54










              • No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
                – Alberto Torrejon Valenzuela
                Nov 22 at 4:56










              • Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
                – Franco Cabrera
                Nov 22 at 4:56














              0












              0








              0






              You should multiply by the conjugate of the expression, then proceed.



              begin{align}
              frac{left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}\
              end{align}
              begin{align}
              frac{(x^4 -x^3+1)-(x^4+15x^2-5)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}
              end{align}



              Operate and calculate the limit.






              share|cite|improve this answer














              You should multiply by the conjugate of the expression, then proceed.



              begin{align}
              frac{left(sqrt{x^4 -x^3+1}-sqrt{x^4+15x^2-5},right)left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}\
              end{align}
              begin{align}
              frac{(x^4 -x^3+1)-(x^4+15x^2-5)}{left(sqrt{x^4 -x^3+1}+sqrt{x^4+15x^2-5},right)}
              end{align}



              Operate and calculate the limit.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 22 at 6:03









              b00n heT

              10.2k12134




              10.2k12134










              answered Nov 22 at 4:48









              Alberto Torrejon Valenzuela

              276




              276












              • negative! that's cool!
                – Christopher Marley
                Nov 22 at 4:52










              • What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
                – Alberto Torrejon Valenzuela
                Nov 22 at 4:54










              • and then, should i evaluate de sign of de function to see the lateral limits?
                – Franco Cabrera
                Nov 22 at 4:54










              • No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
                – Alberto Torrejon Valenzuela
                Nov 22 at 4:56










              • Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
                – Franco Cabrera
                Nov 22 at 4:56


















              • negative! that's cool!
                – Christopher Marley
                Nov 22 at 4:52










              • What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
                – Alberto Torrejon Valenzuela
                Nov 22 at 4:54










              • and then, should i evaluate de sign of de function to see the lateral limits?
                – Franco Cabrera
                Nov 22 at 4:54










              • No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
                – Alberto Torrejon Valenzuela
                Nov 22 at 4:56










              • Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
                – Franco Cabrera
                Nov 22 at 4:56
















              negative! that's cool!
              – Christopher Marley
              Nov 22 at 4:52




              negative! that's cool!
              – Christopher Marley
              Nov 22 at 4:52












              What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
              – Alberto Torrejon Valenzuela
              Nov 22 at 4:54




              What does that mean? Sorry I`m not a fully native english speaker so those two words in the same sentence make me confuse.
              – Alberto Torrejon Valenzuela
              Nov 22 at 4:54












              and then, should i evaluate de sign of de function to see the lateral limits?
              – Franco Cabrera
              Nov 22 at 4:54




              and then, should i evaluate de sign of de function to see the lateral limits?
              – Franco Cabrera
              Nov 22 at 4:54












              No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
              – Alberto Torrejon Valenzuela
              Nov 22 at 4:56




              No need if you just want to give a result of the limit. If you want to study the function then I suppose you should
              – Alberto Torrejon Valenzuela
              Nov 22 at 4:56












              Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
              – Franco Cabrera
              Nov 22 at 4:56




              Alberto, despues de eso tendria que evaluar el signo de f? ver si los limites laterales son iguales y obtener su valor?
              – Franco Cabrera
              Nov 22 at 4:56











              -1














              Another manipulation:



              Let $a,b >0$, real.



              $a-b=(√a-√b)(√a+√b)$, then



              $√a-√b = dfrac{a-b}{√a+√b}.$



              $a: = (x^4-x^3+1)^{1/2},$
              $b=(x^4+15x^2-5)^{1/2}.$



              $small{f(x):=}$



              $small {=dfrac{-x^3-15x^2+6}{(x^4-x^3+1)^{1/2}+(x^4+15x^2-5)^{1/2}}}$



              Note : For large x (positive) the numerator is negative, the denominator positive: Increasing the denominator makes the fraction bigger.



              Hence



              $f(x) lt $



              $dfrac{-x^3+6}{(x^4+1)^{1/2}+(x^4+15x^2)^{1/2}} lt $



              $dfrac{-x^3+6}{√2x^2+√2x^2}=$



              $dfrac{-x^3+6}{(2√2)x^2}.$



              Take the limit.






              share|cite|improve this answer




























                -1














                Another manipulation:



                Let $a,b >0$, real.



                $a-b=(√a-√b)(√a+√b)$, then



                $√a-√b = dfrac{a-b}{√a+√b}.$



                $a: = (x^4-x^3+1)^{1/2},$
                $b=(x^4+15x^2-5)^{1/2}.$



                $small{f(x):=}$



                $small {=dfrac{-x^3-15x^2+6}{(x^4-x^3+1)^{1/2}+(x^4+15x^2-5)^{1/2}}}$



                Note : For large x (positive) the numerator is negative, the denominator positive: Increasing the denominator makes the fraction bigger.



                Hence



                $f(x) lt $



                $dfrac{-x^3+6}{(x^4+1)^{1/2}+(x^4+15x^2)^{1/2}} lt $



                $dfrac{-x^3+6}{√2x^2+√2x^2}=$



                $dfrac{-x^3+6}{(2√2)x^2}.$



                Take the limit.






                share|cite|improve this answer


























                  -1












                  -1








                  -1






                  Another manipulation:



                  Let $a,b >0$, real.



                  $a-b=(√a-√b)(√a+√b)$, then



                  $√a-√b = dfrac{a-b}{√a+√b}.$



                  $a: = (x^4-x^3+1)^{1/2},$
                  $b=(x^4+15x^2-5)^{1/2}.$



                  $small{f(x):=}$



                  $small {=dfrac{-x^3-15x^2+6}{(x^4-x^3+1)^{1/2}+(x^4+15x^2-5)^{1/2}}}$



                  Note : For large x (positive) the numerator is negative, the denominator positive: Increasing the denominator makes the fraction bigger.



                  Hence



                  $f(x) lt $



                  $dfrac{-x^3+6}{(x^4+1)^{1/2}+(x^4+15x^2)^{1/2}} lt $



                  $dfrac{-x^3+6}{√2x^2+√2x^2}=$



                  $dfrac{-x^3+6}{(2√2)x^2}.$



                  Take the limit.






                  share|cite|improve this answer














                  Another manipulation:



                  Let $a,b >0$, real.



                  $a-b=(√a-√b)(√a+√b)$, then



                  $√a-√b = dfrac{a-b}{√a+√b}.$



                  $a: = (x^4-x^3+1)^{1/2},$
                  $b=(x^4+15x^2-5)^{1/2}.$



                  $small{f(x):=}$



                  $small {=dfrac{-x^3-15x^2+6}{(x^4-x^3+1)^{1/2}+(x^4+15x^2-5)^{1/2}}}$



                  Note : For large x (positive) the numerator is negative, the denominator positive: Increasing the denominator makes the fraction bigger.



                  Hence



                  $f(x) lt $



                  $dfrac{-x^3+6}{(x^4+1)^{1/2}+(x^4+15x^2)^{1/2}} lt $



                  $dfrac{-x^3+6}{√2x^2+√2x^2}=$



                  $dfrac{-x^3+6}{(2√2)x^2}.$



                  Take the limit.







                  share|cite|improve this answer














                  share|cite|improve this answer



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                  edited Nov 22 at 9:44

























                  answered Nov 22 at 9:34









                  Peter Szilas

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