For what $p,q,r$ $lim_{x to 0} f(x)=frac{p + qcos x + rsin x}{x^2}=1/2$?












3














Let $$f(x)=frac{p + qcos x + rsin x}{x^2}$$. Then for what values of $p$,$q$ and $r$ is the limit $$lim_{x to 0} f(x)=1/2.$$ I’ve tried using L’Hospital’s rule but couldn’t get anywhere. Any help would be appreciated.






real-analysis sequences-and-series limits







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  • Break into fractions and then you can see what you can do.
    – jayant98
    Dec 7 at 14:54






  • 1




    Can you use Taylor expansions?
    – MisterRiemann
    Dec 7 at 14:59






  • 2




    I believe you are missing a term in the numerator.
    – user10354138
    Dec 7 at 15:01






  • 1




    Indeed, right now this doesn't seem to have a solution.
    – RcnSc
    Dec 7 at 15:03








  • 2




    Hint (as @MisterRiemann suggests). Write out the first few terms of the power series for $sin$ and $cos$, do the algebra and see what you can conclude. L'Hopital is almost always a bad first tool for a job like this.
    – Ethan Bolker
    Dec 7 at 15:12
















3














Let $$f(x)=frac{p + qcos x + rsin x}{x^2}$$. Then for what values of $p$,$q$ and $r$ is the limit $$lim_{x to 0} f(x)=1/2.$$ I’ve tried using L’Hospital’s rule but couldn’t get anywhere. Any help would be appreciated.






real-analysis sequences-and-series limits







share|cite|improve this question
























  • Break into fractions and then you can see what you can do.
    – jayant98
    Dec 7 at 14:54






  • 1




    Can you use Taylor expansions?
    – MisterRiemann
    Dec 7 at 14:59






  • 2




    I believe you are missing a term in the numerator.
    – user10354138
    Dec 7 at 15:01






  • 1




    Indeed, right now this doesn't seem to have a solution.
    – RcnSc
    Dec 7 at 15:03








  • 2




    Hint (as @MisterRiemann suggests). Write out the first few terms of the power series for $sin$ and $cos$, do the algebra and see what you can conclude. L'Hopital is almost always a bad first tool for a job like this.
    – Ethan Bolker
    Dec 7 at 15:12














3












3








3


1





Let $$f(x)=frac{p + qcos x + rsin x}{x^2}$$. Then for what values of $p$,$q$ and $r$ is the limit $$lim_{x to 0} f(x)=1/2.$$ I’ve tried using L’Hospital’s rule but couldn’t get anywhere. Any help would be appreciated.






real-analysis sequences-and-series limits







share|cite|improve this question















Let $$f(x)=frac{p + qcos x + rsin x}{x^2}$$. Then for what values of $p$,$q$ and $r$ is the limit $$lim_{x to 0} f(x)=1/2.$$ I’ve tried using L’Hospital’s rule but couldn’t get anywhere. Any help would be appreciated.






real-analysis sequences-and-series limits




real-analysis sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 at 23:58









Andrews

354317




354317










asked Dec 7 at 14:51









John Mitchell

376210




376210












  • Break into fractions and then you can see what you can do.
    – jayant98
    Dec 7 at 14:54






  • 1




    Can you use Taylor expansions?
    – MisterRiemann
    Dec 7 at 14:59






  • 2




    I believe you are missing a term in the numerator.
    – user10354138
    Dec 7 at 15:01






  • 1




    Indeed, right now this doesn't seem to have a solution.
    – RcnSc
    Dec 7 at 15:03








  • 2




    Hint (as @MisterRiemann suggests). Write out the first few terms of the power series for $sin$ and $cos$, do the algebra and see what you can conclude. L'Hopital is almost always a bad first tool for a job like this.
    – Ethan Bolker
    Dec 7 at 15:12


















  • Break into fractions and then you can see what you can do.
    – jayant98
    Dec 7 at 14:54






  • 1




    Can you use Taylor expansions?
    – MisterRiemann
    Dec 7 at 14:59






  • 2




    I believe you are missing a term in the numerator.
    – user10354138
    Dec 7 at 15:01






  • 1




    Indeed, right now this doesn't seem to have a solution.
    – RcnSc
    Dec 7 at 15:03








  • 2




    Hint (as @MisterRiemann suggests). Write out the first few terms of the power series for $sin$ and $cos$, do the algebra and see what you can conclude. L'Hopital is almost always a bad first tool for a job like this.
    – Ethan Bolker
    Dec 7 at 15:12
















Break into fractions and then you can see what you can do.
– jayant98
Dec 7 at 14:54




Break into fractions and then you can see what you can do.
– jayant98
Dec 7 at 14:54




1




1




Can you use Taylor expansions?
– MisterRiemann
Dec 7 at 14:59




Can you use Taylor expansions?
– MisterRiemann
Dec 7 at 14:59




2




2




I believe you are missing a term in the numerator.
– user10354138
Dec 7 at 15:01




I believe you are missing a term in the numerator.
– user10354138
Dec 7 at 15:01




1




1




Indeed, right now this doesn't seem to have a solution.
– RcnSc
Dec 7 at 15:03






Indeed, right now this doesn't seem to have a solution.
– RcnSc
Dec 7 at 15:03






2




2




Hint (as @MisterRiemann suggests). Write out the first few terms of the power series for $sin$ and $cos$, do the algebra and see what you can conclude. L'Hopital is almost always a bad first tool for a job like this.
– Ethan Bolker
Dec 7 at 15:12




Hint (as @MisterRiemann suggests). Write out the first few terms of the power series for $sin$ and $cos$, do the algebra and see what you can conclude. L'Hopital is almost always a bad first tool for a job like this.
– Ethan Bolker
Dec 7 at 15:12










3 Answers
3






active

oldest

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5














No calculus is needed. Since $frac{sin x}{x}to 1$, $frac{cos x-1}{x^2}=-frac{2sin^2 x/2}{x^2}to-frac{1}{2}$. But $frac{1}{x^2},,frac{sin x}{x^2}$ diverge, so take $p=1,,q=-1,,r=0$.






share|cite|improve this answer

















  • 1




    Well, you're using standard limits, so I don't think it's fair to say that no calculus is needed. :-) Nice solution otherwise (+1).
    – MisterRiemann
    Dec 7 at 15:23








  • 2




    @MisterRiemann No, these limits are famously provable without calculus. I just proved one from the other, whereas $frac{sin x}{x}$ is the portion of an angle-$x$ sector contained in the triangle with the same vertices. (That's not the whole proof, but I'll leave you to look up the rest.)
    – J.G.
    Dec 7 at 15:24












  • Good point! Seems that I have forgotten how these standard limits are actually proven!
    – MisterRiemann
    Dec 7 at 15:26










  • I think the right term should be "no derivatives". I consider limits to be very much a part of calculus. +1
    – Paramanand Singh
    Dec 8 at 2:17












  • @ParamanandSingh I see them more as analysis.
    – J.G.
    Dec 8 at 7:43





















1














$$f(x)=frac{p + qcos x + rsin x}{x^2}$$.



$$lim_{xrightarrow 0}f(x)=frac{1}{2} implies p+q=0$$



$$ lim_{xrightarrow 0}f(x)= lim_{xrightarrow 0} frac{-qsin x+rcos x}{2x}=frac{1}{2} implies r=0$$



$$lim_{xrightarrow 0}f(x)=lim_{xrightarrow 0}frac{-qcos x -rsin x}{2}=frac{1}{2}implies -q=1 implies q=-1.$$



Therefore, $p=1, p=-1,$ and $r=0$.






share|cite|improve this answer





















  • The use of L'Hospital's Rule in reverse is not allowed. You may want to fix it.
    – Paramanand Singh
    Dec 8 at 2:18



















0














We are given that $$lim_{xto 0}frac{p+qcos x+rsin x} {x^2}=frac{1}{2}tag{1}$$ so that multiplication by $x^2$ gives $$lim_{xto 0}(p+qcos x+rsin x) =lim_{xto 0}frac{p+qcos x+rsin x} {x^2}cdot x^2=frac{1}{2}cdot 0=0$$ and hence $p+q=0$.



Next note that $$p+qcos x+rsin x=p+q-q(1-cos x) +rsin x=rsin x-q(1-cos x) $$ and since $(1-cos x) /x^2to 1/2$ it follows from $(1)$ that $$lim_{xto 0}frac{rsin x} {x^2}=frac{1+q}{2}$$ and since $(sin x) /xto 1$ we have $$lim_{xto 0}frac{r}{x}=frac{1+q}{2}tag{2}$$ Multiplying by $x$ gives us $r=0$ and then from the above equation we get $1+q=0$ so that $p=1,q=-1,r=0$.






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    3 Answers
    3






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    3 Answers
    3






    active

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    oldest

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    active

    oldest

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    5














    No calculus is needed. Since $frac{sin x}{x}to 1$, $frac{cos x-1}{x^2}=-frac{2sin^2 x/2}{x^2}to-frac{1}{2}$. But $frac{1}{x^2},,frac{sin x}{x^2}$ diverge, so take $p=1,,q=-1,,r=0$.






    share|cite|improve this answer

















    • 1




      Well, you're using standard limits, so I don't think it's fair to say that no calculus is needed. :-) Nice solution otherwise (+1).
      – MisterRiemann
      Dec 7 at 15:23








    • 2




      @MisterRiemann No, these limits are famously provable without calculus. I just proved one from the other, whereas $frac{sin x}{x}$ is the portion of an angle-$x$ sector contained in the triangle with the same vertices. (That's not the whole proof, but I'll leave you to look up the rest.)
      – J.G.
      Dec 7 at 15:24












    • Good point! Seems that I have forgotten how these standard limits are actually proven!
      – MisterRiemann
      Dec 7 at 15:26










    • I think the right term should be "no derivatives". I consider limits to be very much a part of calculus. +1
      – Paramanand Singh
      Dec 8 at 2:17












    • @ParamanandSingh I see them more as analysis.
      – J.G.
      Dec 8 at 7:43


















    5














    No calculus is needed. Since $frac{sin x}{x}to 1$, $frac{cos x-1}{x^2}=-frac{2sin^2 x/2}{x^2}to-frac{1}{2}$. But $frac{1}{x^2},,frac{sin x}{x^2}$ diverge, so take $p=1,,q=-1,,r=0$.






    share|cite|improve this answer

















    • 1




      Well, you're using standard limits, so I don't think it's fair to say that no calculus is needed. :-) Nice solution otherwise (+1).
      – MisterRiemann
      Dec 7 at 15:23








    • 2




      @MisterRiemann No, these limits are famously provable without calculus. I just proved one from the other, whereas $frac{sin x}{x}$ is the portion of an angle-$x$ sector contained in the triangle with the same vertices. (That's not the whole proof, but I'll leave you to look up the rest.)
      – J.G.
      Dec 7 at 15:24












    • Good point! Seems that I have forgotten how these standard limits are actually proven!
      – MisterRiemann
      Dec 7 at 15:26










    • I think the right term should be "no derivatives". I consider limits to be very much a part of calculus. +1
      – Paramanand Singh
      Dec 8 at 2:17












    • @ParamanandSingh I see them more as analysis.
      – J.G.
      Dec 8 at 7:43
















    5












    5








    5






    No calculus is needed. Since $frac{sin x}{x}to 1$, $frac{cos x-1}{x^2}=-frac{2sin^2 x/2}{x^2}to-frac{1}{2}$. But $frac{1}{x^2},,frac{sin x}{x^2}$ diverge, so take $p=1,,q=-1,,r=0$.






    share|cite|improve this answer












    No calculus is needed. Since $frac{sin x}{x}to 1$, $frac{cos x-1}{x^2}=-frac{2sin^2 x/2}{x^2}to-frac{1}{2}$. But $frac{1}{x^2},,frac{sin x}{x^2}$ diverge, so take $p=1,,q=-1,,r=0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 7 at 15:14









    J.G.

    22.1k22034




    22.1k22034








    • 1




      Well, you're using standard limits, so I don't think it's fair to say that no calculus is needed. :-) Nice solution otherwise (+1).
      – MisterRiemann
      Dec 7 at 15:23








    • 2




      @MisterRiemann No, these limits are famously provable without calculus. I just proved one from the other, whereas $frac{sin x}{x}$ is the portion of an angle-$x$ sector contained in the triangle with the same vertices. (That's not the whole proof, but I'll leave you to look up the rest.)
      – J.G.
      Dec 7 at 15:24












    • Good point! Seems that I have forgotten how these standard limits are actually proven!
      – MisterRiemann
      Dec 7 at 15:26










    • I think the right term should be "no derivatives". I consider limits to be very much a part of calculus. +1
      – Paramanand Singh
      Dec 8 at 2:17












    • @ParamanandSingh I see them more as analysis.
      – J.G.
      Dec 8 at 7:43
















    • 1




      Well, you're using standard limits, so I don't think it's fair to say that no calculus is needed. :-) Nice solution otherwise (+1).
      – MisterRiemann
      Dec 7 at 15:23








    • 2




      @MisterRiemann No, these limits are famously provable without calculus. I just proved one from the other, whereas $frac{sin x}{x}$ is the portion of an angle-$x$ sector contained in the triangle with the same vertices. (That's not the whole proof, but I'll leave you to look up the rest.)
      – J.G.
      Dec 7 at 15:24












    • Good point! Seems that I have forgotten how these standard limits are actually proven!
      – MisterRiemann
      Dec 7 at 15:26










    • I think the right term should be "no derivatives". I consider limits to be very much a part of calculus. +1
      – Paramanand Singh
      Dec 8 at 2:17












    • @ParamanandSingh I see them more as analysis.
      – J.G.
      Dec 8 at 7:43










    1




    1




    Well, you're using standard limits, so I don't think it's fair to say that no calculus is needed. :-) Nice solution otherwise (+1).
    – MisterRiemann
    Dec 7 at 15:23






    Well, you're using standard limits, so I don't think it's fair to say that no calculus is needed. :-) Nice solution otherwise (+1).
    – MisterRiemann
    Dec 7 at 15:23






    2




    2




    @MisterRiemann No, these limits are famously provable without calculus. I just proved one from the other, whereas $frac{sin x}{x}$ is the portion of an angle-$x$ sector contained in the triangle with the same vertices. (That's not the whole proof, but I'll leave you to look up the rest.)
    – J.G.
    Dec 7 at 15:24






    @MisterRiemann No, these limits are famously provable without calculus. I just proved one from the other, whereas $frac{sin x}{x}$ is the portion of an angle-$x$ sector contained in the triangle with the same vertices. (That's not the whole proof, but I'll leave you to look up the rest.)
    – J.G.
    Dec 7 at 15:24














    Good point! Seems that I have forgotten how these standard limits are actually proven!
    – MisterRiemann
    Dec 7 at 15:26




    Good point! Seems that I have forgotten how these standard limits are actually proven!
    – MisterRiemann
    Dec 7 at 15:26












    I think the right term should be "no derivatives". I consider limits to be very much a part of calculus. +1
    – Paramanand Singh
    Dec 8 at 2:17






    I think the right term should be "no derivatives". I consider limits to be very much a part of calculus. +1
    – Paramanand Singh
    Dec 8 at 2:17














    @ParamanandSingh I see them more as analysis.
    – J.G.
    Dec 8 at 7:43






    @ParamanandSingh I see them more as analysis.
    – J.G.
    Dec 8 at 7:43













    1














    $$f(x)=frac{p + qcos x + rsin x}{x^2}$$.



    $$lim_{xrightarrow 0}f(x)=frac{1}{2} implies p+q=0$$



    $$ lim_{xrightarrow 0}f(x)= lim_{xrightarrow 0} frac{-qsin x+rcos x}{2x}=frac{1}{2} implies r=0$$



    $$lim_{xrightarrow 0}f(x)=lim_{xrightarrow 0}frac{-qcos x -rsin x}{2}=frac{1}{2}implies -q=1 implies q=-1.$$



    Therefore, $p=1, p=-1,$ and $r=0$.






    share|cite|improve this answer





















    • The use of L'Hospital's Rule in reverse is not allowed. You may want to fix it.
      – Paramanand Singh
      Dec 8 at 2:18
















    1














    $$f(x)=frac{p + qcos x + rsin x}{x^2}$$.



    $$lim_{xrightarrow 0}f(x)=frac{1}{2} implies p+q=0$$



    $$ lim_{xrightarrow 0}f(x)= lim_{xrightarrow 0} frac{-qsin x+rcos x}{2x}=frac{1}{2} implies r=0$$



    $$lim_{xrightarrow 0}f(x)=lim_{xrightarrow 0}frac{-qcos x -rsin x}{2}=frac{1}{2}implies -q=1 implies q=-1.$$



    Therefore, $p=1, p=-1,$ and $r=0$.






    share|cite|improve this answer





















    • The use of L'Hospital's Rule in reverse is not allowed. You may want to fix it.
      – Paramanand Singh
      Dec 8 at 2:18














    1












    1








    1






    $$f(x)=frac{p + qcos x + rsin x}{x^2}$$.



    $$lim_{xrightarrow 0}f(x)=frac{1}{2} implies p+q=0$$



    $$ lim_{xrightarrow 0}f(x)= lim_{xrightarrow 0} frac{-qsin x+rcos x}{2x}=frac{1}{2} implies r=0$$



    $$lim_{xrightarrow 0}f(x)=lim_{xrightarrow 0}frac{-qcos x -rsin x}{2}=frac{1}{2}implies -q=1 implies q=-1.$$



    Therefore, $p=1, p=-1,$ and $r=0$.






    share|cite|improve this answer












    $$f(x)=frac{p + qcos x + rsin x}{x^2}$$.



    $$lim_{xrightarrow 0}f(x)=frac{1}{2} implies p+q=0$$



    $$ lim_{xrightarrow 0}f(x)= lim_{xrightarrow 0} frac{-qsin x+rcos x}{2x}=frac{1}{2} implies r=0$$



    $$lim_{xrightarrow 0}f(x)=lim_{xrightarrow 0}frac{-qcos x -rsin x}{2}=frac{1}{2}implies -q=1 implies q=-1.$$



    Therefore, $p=1, p=-1,$ and $r=0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 7 at 15:16









    LeB

    979217




    979217












    • The use of L'Hospital's Rule in reverse is not allowed. You may want to fix it.
      – Paramanand Singh
      Dec 8 at 2:18


















    • The use of L'Hospital's Rule in reverse is not allowed. You may want to fix it.
      – Paramanand Singh
      Dec 8 at 2:18
















    The use of L'Hospital's Rule in reverse is not allowed. You may want to fix it.
    – Paramanand Singh
    Dec 8 at 2:18




    The use of L'Hospital's Rule in reverse is not allowed. You may want to fix it.
    – Paramanand Singh
    Dec 8 at 2:18











    0














    We are given that $$lim_{xto 0}frac{p+qcos x+rsin x} {x^2}=frac{1}{2}tag{1}$$ so that multiplication by $x^2$ gives $$lim_{xto 0}(p+qcos x+rsin x) =lim_{xto 0}frac{p+qcos x+rsin x} {x^2}cdot x^2=frac{1}{2}cdot 0=0$$ and hence $p+q=0$.



    Next note that $$p+qcos x+rsin x=p+q-q(1-cos x) +rsin x=rsin x-q(1-cos x) $$ and since $(1-cos x) /x^2to 1/2$ it follows from $(1)$ that $$lim_{xto 0}frac{rsin x} {x^2}=frac{1+q}{2}$$ and since $(sin x) /xto 1$ we have $$lim_{xto 0}frac{r}{x}=frac{1+q}{2}tag{2}$$ Multiplying by $x$ gives us $r=0$ and then from the above equation we get $1+q=0$ so that $p=1,q=-1,r=0$.






    share|cite|improve this answer


























      0














      We are given that $$lim_{xto 0}frac{p+qcos x+rsin x} {x^2}=frac{1}{2}tag{1}$$ so that multiplication by $x^2$ gives $$lim_{xto 0}(p+qcos x+rsin x) =lim_{xto 0}frac{p+qcos x+rsin x} {x^2}cdot x^2=frac{1}{2}cdot 0=0$$ and hence $p+q=0$.



      Next note that $$p+qcos x+rsin x=p+q-q(1-cos x) +rsin x=rsin x-q(1-cos x) $$ and since $(1-cos x) /x^2to 1/2$ it follows from $(1)$ that $$lim_{xto 0}frac{rsin x} {x^2}=frac{1+q}{2}$$ and since $(sin x) /xto 1$ we have $$lim_{xto 0}frac{r}{x}=frac{1+q}{2}tag{2}$$ Multiplying by $x$ gives us $r=0$ and then from the above equation we get $1+q=0$ so that $p=1,q=-1,r=0$.






      share|cite|improve this answer
























        0












        0








        0






        We are given that $$lim_{xto 0}frac{p+qcos x+rsin x} {x^2}=frac{1}{2}tag{1}$$ so that multiplication by $x^2$ gives $$lim_{xto 0}(p+qcos x+rsin x) =lim_{xto 0}frac{p+qcos x+rsin x} {x^2}cdot x^2=frac{1}{2}cdot 0=0$$ and hence $p+q=0$.



        Next note that $$p+qcos x+rsin x=p+q-q(1-cos x) +rsin x=rsin x-q(1-cos x) $$ and since $(1-cos x) /x^2to 1/2$ it follows from $(1)$ that $$lim_{xto 0}frac{rsin x} {x^2}=frac{1+q}{2}$$ and since $(sin x) /xto 1$ we have $$lim_{xto 0}frac{r}{x}=frac{1+q}{2}tag{2}$$ Multiplying by $x$ gives us $r=0$ and then from the above equation we get $1+q=0$ so that $p=1,q=-1,r=0$.






        share|cite|improve this answer












        We are given that $$lim_{xto 0}frac{p+qcos x+rsin x} {x^2}=frac{1}{2}tag{1}$$ so that multiplication by $x^2$ gives $$lim_{xto 0}(p+qcos x+rsin x) =lim_{xto 0}frac{p+qcos x+rsin x} {x^2}cdot x^2=frac{1}{2}cdot 0=0$$ and hence $p+q=0$.



        Next note that $$p+qcos x+rsin x=p+q-q(1-cos x) +rsin x=rsin x-q(1-cos x) $$ and since $(1-cos x) /x^2to 1/2$ it follows from $(1)$ that $$lim_{xto 0}frac{rsin x} {x^2}=frac{1+q}{2}$$ and since $(sin x) /xto 1$ we have $$lim_{xto 0}frac{r}{x}=frac{1+q}{2}tag{2}$$ Multiplying by $x$ gives us $r=0$ and then from the above equation we get $1+q=0$ so that $p=1,q=-1,r=0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 at 2:32









        Paramanand Singh

        48.8k555157




        48.8k555157






























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