Given $f(x)=o(1), g(x)=O(h(x)) text{ as } xto a $, show that $f(x)g(x)=o(h(x))$












1












$begingroup$


We are given that $f(x)=o(1)$ as $ xto a $, $g(x)=O(h(x))$ as $ xto a $. It is required to prove or show that $f(x)g(x)=o(h(x))$ as $ xto a $



I just multiplied the two functions to get: $$f(x)g(x)=o(1)O(h(x))=color{red}{o(1)O(1)h(x) = o(1)h(x)=o(h(x))} text{ as $xto a$ } $$



My main question is whether or not the part in red is a valid use of the properties of asymptotic expressions.






calculus limits discrete-mathematics asymptotics







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    We are given that $f(x)=o(1)$ as $ xto a $, $g(x)=O(h(x))$ as $ xto a $. It is required to prove or show that $f(x)g(x)=o(h(x))$ as $ xto a $



    I just multiplied the two functions to get: $$f(x)g(x)=o(1)O(h(x))=color{red}{o(1)O(1)h(x) = o(1)h(x)=o(h(x))} text{ as $xto a$ } $$



    My main question is whether or not the part in red is a valid use of the properties of asymptotic expressions.






    calculus limits discrete-mathematics asymptotics







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      We are given that $f(x)=o(1)$ as $ xto a $, $g(x)=O(h(x))$ as $ xto a $. It is required to prove or show that $f(x)g(x)=o(h(x))$ as $ xto a $



      I just multiplied the two functions to get: $$f(x)g(x)=o(1)O(h(x))=color{red}{o(1)O(1)h(x) = o(1)h(x)=o(h(x))} text{ as $xto a$ } $$



      My main question is whether or not the part in red is a valid use of the properties of asymptotic expressions.






      calculus limits discrete-mathematics asymptotics







      share|cite|improve this question









      $endgroup$




      We are given that $f(x)=o(1)$ as $ xto a $, $g(x)=O(h(x))$ as $ xto a $. It is required to prove or show that $f(x)g(x)=o(h(x))$ as $ xto a $



      I just multiplied the two functions to get: $$f(x)g(x)=o(1)O(h(x))=color{red}{o(1)O(1)h(x) = o(1)h(x)=o(h(x))} text{ as $xto a$ } $$



      My main question is whether or not the part in red is a valid use of the properties of asymptotic expressions.






      calculus limits discrete-mathematics asymptotics




      calculus limits discrete-mathematics asymptotics






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 14 '18 at 10:27









      E.NoleE.Nole

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      113112






















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          1












          $begingroup$

          Recall that by definition as $xto a$



          $$f(x)=o(1)iff f(x)=1cdot omega_1(x) quad omega_1(x)to 0$$



          $$g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$$



          then



          $$f(x)g(x)=h(x)cdotomega_1(x)cdotomega_2(x)=h(x)cdot omega_3(x) quad omega_3(x)to 0$$



          indeed we have that $|omega_2(x)|le M$ and by squeeze theorem



          $$|omega_3(x)|=|omega_1(x)cdotomega_2(x)|le M|omega_1(x)to 0$$



          In the same way we can see that




          • $O(h(x))=h(x)cdot O(1)$


          • $o(1)cdot O(1)=o(1)$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
            $endgroup$
            – E.Nole
            Dec 14 '18 at 15:40








          • 1




            $begingroup$
            Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
            $endgroup$
            – gimusi
            Dec 14 '18 at 15:50











          Your Answer





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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          Recall that by definition as $xto a$



          $$f(x)=o(1)iff f(x)=1cdot omega_1(x) quad omega_1(x)to 0$$



          $$g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$$



          then



          $$f(x)g(x)=h(x)cdotomega_1(x)cdotomega_2(x)=h(x)cdot omega_3(x) quad omega_3(x)to 0$$



          indeed we have that $|omega_2(x)|le M$ and by squeeze theorem



          $$|omega_3(x)|=|omega_1(x)cdotomega_2(x)|le M|omega_1(x)to 0$$



          In the same way we can see that




          • $O(h(x))=h(x)cdot O(1)$


          • $o(1)cdot O(1)=o(1)$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
            $endgroup$
            – E.Nole
            Dec 14 '18 at 15:40








          • 1




            $begingroup$
            Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
            $endgroup$
            – gimusi
            Dec 14 '18 at 15:50
















          1












          $begingroup$

          Recall that by definition as $xto a$



          $$f(x)=o(1)iff f(x)=1cdot omega_1(x) quad omega_1(x)to 0$$



          $$g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$$



          then



          $$f(x)g(x)=h(x)cdotomega_1(x)cdotomega_2(x)=h(x)cdot omega_3(x) quad omega_3(x)to 0$$



          indeed we have that $|omega_2(x)|le M$ and by squeeze theorem



          $$|omega_3(x)|=|omega_1(x)cdotomega_2(x)|le M|omega_1(x)to 0$$



          In the same way we can see that




          • $O(h(x))=h(x)cdot O(1)$


          • $o(1)cdot O(1)=o(1)$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
            $endgroup$
            – E.Nole
            Dec 14 '18 at 15:40








          • 1




            $begingroup$
            Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
            $endgroup$
            – gimusi
            Dec 14 '18 at 15:50














          1












          1








          1





          $begingroup$

          Recall that by definition as $xto a$



          $$f(x)=o(1)iff f(x)=1cdot omega_1(x) quad omega_1(x)to 0$$



          $$g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$$



          then



          $$f(x)g(x)=h(x)cdotomega_1(x)cdotomega_2(x)=h(x)cdot omega_3(x) quad omega_3(x)to 0$$



          indeed we have that $|omega_2(x)|le M$ and by squeeze theorem



          $$|omega_3(x)|=|omega_1(x)cdotomega_2(x)|le M|omega_1(x)to 0$$



          In the same way we can see that




          • $O(h(x))=h(x)cdot O(1)$


          • $o(1)cdot O(1)=o(1)$







          share|cite|improve this answer











          $endgroup$



          Recall that by definition as $xto a$



          $$f(x)=o(1)iff f(x)=1cdot omega_1(x) quad omega_1(x)to 0$$



          $$g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$$



          then



          $$f(x)g(x)=h(x)cdotomega_1(x)cdotomega_2(x)=h(x)cdot omega_3(x) quad omega_3(x)to 0$$



          indeed we have that $|omega_2(x)|le M$ and by squeeze theorem



          $$|omega_3(x)|=|omega_1(x)cdotomega_2(x)|le M|omega_1(x)to 0$$



          In the same way we can see that




          • $O(h(x))=h(x)cdot O(1)$


          • $o(1)cdot O(1)=o(1)$








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 10:45

























          answered Dec 14 '18 at 10:30









          gimusigimusi

          1




          1












          • $begingroup$
            I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
            $endgroup$
            – E.Nole
            Dec 14 '18 at 15:40








          • 1




            $begingroup$
            Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
            $endgroup$
            – gimusi
            Dec 14 '18 at 15:50


















          • $begingroup$
            I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
            $endgroup$
            – E.Nole
            Dec 14 '18 at 15:40








          • 1




            $begingroup$
            Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
            $endgroup$
            – gimusi
            Dec 14 '18 at 15:50
















          $begingroup$
          I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
          $endgroup$
          – E.Nole
          Dec 14 '18 at 15:40






          $begingroup$
          I was given slightly different definitions so I'd like to clarify, How is $g(x)=O(h(x)) iff g(x)=h(x)cdot omega_2(x) quad limsup|omega_2(x)|in mathbb R$? The limsup part is where I'm particularly lost. The definition I know is that $g(x)=O(h(x)) iff lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$
          $endgroup$
          – E.Nole
          Dec 14 '18 at 15:40






          1




          1




          $begingroup$
          Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
          $endgroup$
          – gimusi
          Dec 14 '18 at 15:50




          $begingroup$
          Your definition $$lim_{xrightarrow a} frac{g(x)}{h(x)} = C < infty$$ is uncorrect we should have $$limsup_{xrightarrow a} left|frac{g(x)}{h(x)}right| = C < infty$$ Refer to formal definition. The definition I used is more general since include also te case with $h(x)=0$.
          $endgroup$
          – gimusi
          Dec 14 '18 at 15:50


















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