What am I doing wrong finding $lim_{xto 0} left( frac{1+xcdot2^x}{1+xcdot3^x} right)^{1/x^2}$?











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It has an answer here, but I'd like to know where my solution went wrong.



$$lim_{xto 0} left( frac{1+xcdot2^x}{1+xcdot3^x} right)^{frac{1}{x^2}} $$
$$lim_{xto 0} left( frac{1+xcdot2^x +xcdot 3^x-xcdot 3^x}{1+xcdot3^x} right)^frac{1}{x^2} $$
$$lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^frac{1}{x^2} $$
$$lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^{frac{1}{x^2}cdot frac{1+xcdot 3^x}{xcdot2^x-xcdot 3^x}cdotfrac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} $$
$$lim_{xto 0} e^{frac{1}{x^2}cdot frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} $$
$$lim_{xto 0} e^{frac{1}{x}cdot frac{2^x-3^x}{1+xcdot3^x}} $$
$$lim_{xto 0} e^{frac{1}{x}cdot frac{(1+1)^x-(2+1)^x}{1+xcdot3^x}} $$
$$lim_{xto 0} e^{frac{1}{x}cdot frac{1+x+o(x)-1-x2-o(x)}{1+xcdot3^x}} $$
$$lim_{xto 0} e^{frac{-1}{1+xcdot3^x}} $$
$$e^{-1}$$
The answer in the book is $frac{2}{3}$.










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    It has an answer here, but I'd like to know where my solution went wrong.



    $$lim_{xto 0} left( frac{1+xcdot2^x}{1+xcdot3^x} right)^{frac{1}{x^2}} $$
    $$lim_{xto 0} left( frac{1+xcdot2^x +xcdot 3^x-xcdot 3^x}{1+xcdot3^x} right)^frac{1}{x^2} $$
    $$lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^frac{1}{x^2} $$
    $$lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^{frac{1}{x^2}cdot frac{1+xcdot 3^x}{xcdot2^x-xcdot 3^x}cdotfrac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} $$
    $$lim_{xto 0} e^{frac{1}{x^2}cdot frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} $$
    $$lim_{xto 0} e^{frac{1}{x}cdot frac{2^x-3^x}{1+xcdot3^x}} $$
    $$lim_{xto 0} e^{frac{1}{x}cdot frac{(1+1)^x-(2+1)^x}{1+xcdot3^x}} $$
    $$lim_{xto 0} e^{frac{1}{x}cdot frac{1+x+o(x)-1-x2-o(x)}{1+xcdot3^x}} $$
    $$lim_{xto 0} e^{frac{-1}{1+xcdot3^x}} $$
    $$e^{-1}$$
    The answer in the book is $frac{2}{3}$.










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      It has an answer here, but I'd like to know where my solution went wrong.



      $$lim_{xto 0} left( frac{1+xcdot2^x}{1+xcdot3^x} right)^{frac{1}{x^2}} $$
      $$lim_{xto 0} left( frac{1+xcdot2^x +xcdot 3^x-xcdot 3^x}{1+xcdot3^x} right)^frac{1}{x^2} $$
      $$lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^frac{1}{x^2} $$
      $$lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^{frac{1}{x^2}cdot frac{1+xcdot 3^x}{xcdot2^x-xcdot 3^x}cdotfrac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} $$
      $$lim_{xto 0} e^{frac{1}{x^2}cdot frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} $$
      $$lim_{xto 0} e^{frac{1}{x}cdot frac{2^x-3^x}{1+xcdot3^x}} $$
      $$lim_{xto 0} e^{frac{1}{x}cdot frac{(1+1)^x-(2+1)^x}{1+xcdot3^x}} $$
      $$lim_{xto 0} e^{frac{1}{x}cdot frac{1+x+o(x)-1-x2-o(x)}{1+xcdot3^x}} $$
      $$lim_{xto 0} e^{frac{-1}{1+xcdot3^x}} $$
      $$e^{-1}$$
      The answer in the book is $frac{2}{3}$.










      share|cite|improve this question















      It has an answer here, but I'd like to know where my solution went wrong.



      $$lim_{xto 0} left( frac{1+xcdot2^x}{1+xcdot3^x} right)^{frac{1}{x^2}} $$
      $$lim_{xto 0} left( frac{1+xcdot2^x +xcdot 3^x-xcdot 3^x}{1+xcdot3^x} right)^frac{1}{x^2} $$
      $$lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^frac{1}{x^2} $$
      $$lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^{frac{1}{x^2}cdot frac{1+xcdot 3^x}{xcdot2^x-xcdot 3^x}cdotfrac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} $$
      $$lim_{xto 0} e^{frac{1}{x^2}cdot frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} $$
      $$lim_{xto 0} e^{frac{1}{x}cdot frac{2^x-3^x}{1+xcdot3^x}} $$
      $$lim_{xto 0} e^{frac{1}{x}cdot frac{(1+1)^x-(2+1)^x}{1+xcdot3^x}} $$
      $$lim_{xto 0} e^{frac{1}{x}cdot frac{1+x+o(x)-1-x2-o(x)}{1+xcdot3^x}} $$
      $$lim_{xto 0} e^{frac{-1}{1+xcdot3^x}} $$
      $$e^{-1}$$
      The answer in the book is $frac{2}{3}$.







      calculus limits limits-without-lhopital






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      edited Nov 18 at 6:32









      StubbornAtom

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      asked Nov 18 at 6:04









      fragileradius

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          You want to evaluate:
          $$lim_{xto 0}frac{1}{x}cdot frac{2^x-3^x}{1+xcdot3^x}=lim_{xto 0}frac{1}{x}cdot frac{(1+xcolor{red}{ln 2}+o(x^2))-(1+xcolor{red}{ln 3}+o(x^2))}{1+xcdot3^x}=\
          lim_{xto 0}frac{ln (2/3)+o(x^2)}{1+xcdot3^x}=ln frac23.$$






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            You have used binomial theorem to expand $2^x$ and $3^x$ and this is not valid as the binomial theorem for a general exponent is valid only under certain conditions. To be precise you have $$(1+t)^x=sum_{k=0}^{infty}binom{x}{k}t^k$$ for $|t|<1$. The result is valid for $t=pm 1$ with certain restrictions on $x$. For details see this blog post.



            But in any case it is not valid for $t=2$ and thus can't be used to expand $3^x$. Also when the binomial expansion is valid we have $$(1+t)^x=1+xt+o(t)$$ and not $$(1+t)^x=1+xt+o(x)$$ The right approach for your problem is to use the standard limit $$lim_{xto 0}frac{a^x-1}{x}=log a$$ and thus we get $$lim_{xto 0}frac{2^x-3^x}{x(1+x3^x)}=lim_{xto 0}frac{2^x-1}{x}-frac{3^x-1}{x}=log 2-log 3=log(2/3)$$ and the desired limit is $e^{log(2/3)}=2/3$.



            Also you can split limit operations into base and exponent under certain conditions but it is much simpler to take logs rather than remembering the conditions under which one can split limit operations.






            share|cite|improve this answer






























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              You went wrong in this step



              $$ldots=lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^{frac{1}{x^2}cdot frac{1+xcdot 3^x}{xcdot2^x-xcdot 3^x}cdotfrac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} =lim_{xto 0} color{red}{e^{frac{1}{x^2}cdot frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}}}= ldots$$



              since you take the limit for a part of the expression.



              Refer also to the related




              • Problem with limit solving

              • Analyzing limits problem Calculus (tell me where I'm wrong).






              share|cite|improve this answer























              • 1. Don't arithmetical properties of limits allow us to take the limit for a part of the expression? 2. Would it had been correct if I have sent the "lim" into the exponent? Thank you!
                – fragileradius
                Nov 18 at 8:08






              • 1




                @fragileradius I've added some references with a full discussion about that.
                – gimusi
                Nov 18 at 8:12











              Your Answer





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              3 Answers
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              You want to evaluate:
              $$lim_{xto 0}frac{1}{x}cdot frac{2^x-3^x}{1+xcdot3^x}=lim_{xto 0}frac{1}{x}cdot frac{(1+xcolor{red}{ln 2}+o(x^2))-(1+xcolor{red}{ln 3}+o(x^2))}{1+xcdot3^x}=\
              lim_{xto 0}frac{ln (2/3)+o(x^2)}{1+xcdot3^x}=ln frac23.$$






              share|cite|improve this answer

























                up vote
                6
                down vote



                accepted










                You want to evaluate:
                $$lim_{xto 0}frac{1}{x}cdot frac{2^x-3^x}{1+xcdot3^x}=lim_{xto 0}frac{1}{x}cdot frac{(1+xcolor{red}{ln 2}+o(x^2))-(1+xcolor{red}{ln 3}+o(x^2))}{1+xcdot3^x}=\
                lim_{xto 0}frac{ln (2/3)+o(x^2)}{1+xcdot3^x}=ln frac23.$$






                share|cite|improve this answer























                  up vote
                  6
                  down vote



                  accepted







                  up vote
                  6
                  down vote



                  accepted






                  You want to evaluate:
                  $$lim_{xto 0}frac{1}{x}cdot frac{2^x-3^x}{1+xcdot3^x}=lim_{xto 0}frac{1}{x}cdot frac{(1+xcolor{red}{ln 2}+o(x^2))-(1+xcolor{red}{ln 3}+o(x^2))}{1+xcdot3^x}=\
                  lim_{xto 0}frac{ln (2/3)+o(x^2)}{1+xcdot3^x}=ln frac23.$$






                  share|cite|improve this answer












                  You want to evaluate:
                  $$lim_{xto 0}frac{1}{x}cdot frac{2^x-3^x}{1+xcdot3^x}=lim_{xto 0}frac{1}{x}cdot frac{(1+xcolor{red}{ln 2}+o(x^2))-(1+xcolor{red}{ln 3}+o(x^2))}{1+xcdot3^x}=\
                  lim_{xto 0}frac{ln (2/3)+o(x^2)}{1+xcdot3^x}=ln frac23.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 18 at 6:21









                  farruhota

                  17.8k2736




                  17.8k2736






















                      up vote
                      3
                      down vote













                      You have used binomial theorem to expand $2^x$ and $3^x$ and this is not valid as the binomial theorem for a general exponent is valid only under certain conditions. To be precise you have $$(1+t)^x=sum_{k=0}^{infty}binom{x}{k}t^k$$ for $|t|<1$. The result is valid for $t=pm 1$ with certain restrictions on $x$. For details see this blog post.



                      But in any case it is not valid for $t=2$ and thus can't be used to expand $3^x$. Also when the binomial expansion is valid we have $$(1+t)^x=1+xt+o(t)$$ and not $$(1+t)^x=1+xt+o(x)$$ The right approach for your problem is to use the standard limit $$lim_{xto 0}frac{a^x-1}{x}=log a$$ and thus we get $$lim_{xto 0}frac{2^x-3^x}{x(1+x3^x)}=lim_{xto 0}frac{2^x-1}{x}-frac{3^x-1}{x}=log 2-log 3=log(2/3)$$ and the desired limit is $e^{log(2/3)}=2/3$.



                      Also you can split limit operations into base and exponent under certain conditions but it is much simpler to take logs rather than remembering the conditions under which one can split limit operations.






                      share|cite|improve this answer



























                        up vote
                        3
                        down vote













                        You have used binomial theorem to expand $2^x$ and $3^x$ and this is not valid as the binomial theorem for a general exponent is valid only under certain conditions. To be precise you have $$(1+t)^x=sum_{k=0}^{infty}binom{x}{k}t^k$$ for $|t|<1$. The result is valid for $t=pm 1$ with certain restrictions on $x$. For details see this blog post.



                        But in any case it is not valid for $t=2$ and thus can't be used to expand $3^x$. Also when the binomial expansion is valid we have $$(1+t)^x=1+xt+o(t)$$ and not $$(1+t)^x=1+xt+o(x)$$ The right approach for your problem is to use the standard limit $$lim_{xto 0}frac{a^x-1}{x}=log a$$ and thus we get $$lim_{xto 0}frac{2^x-3^x}{x(1+x3^x)}=lim_{xto 0}frac{2^x-1}{x}-frac{3^x-1}{x}=log 2-log 3=log(2/3)$$ and the desired limit is $e^{log(2/3)}=2/3$.



                        Also you can split limit operations into base and exponent under certain conditions but it is much simpler to take logs rather than remembering the conditions under which one can split limit operations.






                        share|cite|improve this answer

























                          up vote
                          3
                          down vote










                          up vote
                          3
                          down vote









                          You have used binomial theorem to expand $2^x$ and $3^x$ and this is not valid as the binomial theorem for a general exponent is valid only under certain conditions. To be precise you have $$(1+t)^x=sum_{k=0}^{infty}binom{x}{k}t^k$$ for $|t|<1$. The result is valid for $t=pm 1$ with certain restrictions on $x$. For details see this blog post.



                          But in any case it is not valid for $t=2$ and thus can't be used to expand $3^x$. Also when the binomial expansion is valid we have $$(1+t)^x=1+xt+o(t)$$ and not $$(1+t)^x=1+xt+o(x)$$ The right approach for your problem is to use the standard limit $$lim_{xto 0}frac{a^x-1}{x}=log a$$ and thus we get $$lim_{xto 0}frac{2^x-3^x}{x(1+x3^x)}=lim_{xto 0}frac{2^x-1}{x}-frac{3^x-1}{x}=log 2-log 3=log(2/3)$$ and the desired limit is $e^{log(2/3)}=2/3$.



                          Also you can split limit operations into base and exponent under certain conditions but it is much simpler to take logs rather than remembering the conditions under which one can split limit operations.






                          share|cite|improve this answer














                          You have used binomial theorem to expand $2^x$ and $3^x$ and this is not valid as the binomial theorem for a general exponent is valid only under certain conditions. To be precise you have $$(1+t)^x=sum_{k=0}^{infty}binom{x}{k}t^k$$ for $|t|<1$. The result is valid for $t=pm 1$ with certain restrictions on $x$. For details see this blog post.



                          But in any case it is not valid for $t=2$ and thus can't be used to expand $3^x$. Also when the binomial expansion is valid we have $$(1+t)^x=1+xt+o(t)$$ and not $$(1+t)^x=1+xt+o(x)$$ The right approach for your problem is to use the standard limit $$lim_{xto 0}frac{a^x-1}{x}=log a$$ and thus we get $$lim_{xto 0}frac{2^x-3^x}{x(1+x3^x)}=lim_{xto 0}frac{2^x-1}{x}-frac{3^x-1}{x}=log 2-log 3=log(2/3)$$ and the desired limit is $e^{log(2/3)}=2/3$.



                          Also you can split limit operations into base and exponent under certain conditions but it is much simpler to take logs rather than remembering the conditions under which one can split limit operations.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 18 at 8:27

























                          answered Nov 18 at 8:22









                          Paramanand Singh

                          48.2k555156




                          48.2k555156






















                              up vote
                              2
                              down vote













                              You went wrong in this step



                              $$ldots=lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^{frac{1}{x^2}cdot frac{1+xcdot 3^x}{xcdot2^x-xcdot 3^x}cdotfrac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} =lim_{xto 0} color{red}{e^{frac{1}{x^2}cdot frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}}}= ldots$$



                              since you take the limit for a part of the expression.



                              Refer also to the related




                              • Problem with limit solving

                              • Analyzing limits problem Calculus (tell me where I'm wrong).






                              share|cite|improve this answer























                              • 1. Don't arithmetical properties of limits allow us to take the limit for a part of the expression? 2. Would it had been correct if I have sent the "lim" into the exponent? Thank you!
                                – fragileradius
                                Nov 18 at 8:08






                              • 1




                                @fragileradius I've added some references with a full discussion about that.
                                – gimusi
                                Nov 18 at 8:12















                              up vote
                              2
                              down vote













                              You went wrong in this step



                              $$ldots=lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^{frac{1}{x^2}cdot frac{1+xcdot 3^x}{xcdot2^x-xcdot 3^x}cdotfrac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} =lim_{xto 0} color{red}{e^{frac{1}{x^2}cdot frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}}}= ldots$$



                              since you take the limit for a part of the expression.



                              Refer also to the related




                              • Problem with limit solving

                              • Analyzing limits problem Calculus (tell me where I'm wrong).






                              share|cite|improve this answer























                              • 1. Don't arithmetical properties of limits allow us to take the limit for a part of the expression? 2. Would it had been correct if I have sent the "lim" into the exponent? Thank you!
                                – fragileradius
                                Nov 18 at 8:08






                              • 1




                                @fragileradius I've added some references with a full discussion about that.
                                – gimusi
                                Nov 18 at 8:12













                              up vote
                              2
                              down vote










                              up vote
                              2
                              down vote









                              You went wrong in this step



                              $$ldots=lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^{frac{1}{x^2}cdot frac{1+xcdot 3^x}{xcdot2^x-xcdot 3^x}cdotfrac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} =lim_{xto 0} color{red}{e^{frac{1}{x^2}cdot frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}}}= ldots$$



                              since you take the limit for a part of the expression.



                              Refer also to the related




                              • Problem with limit solving

                              • Analyzing limits problem Calculus (tell me where I'm wrong).






                              share|cite|improve this answer














                              You went wrong in this step



                              $$ldots=lim_{xto 0} left( 1 + frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x} right)^{frac{1}{x^2}cdot frac{1+xcdot 3^x}{xcdot2^x-xcdot 3^x}cdotfrac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}} =lim_{xto 0} color{red}{e^{frac{1}{x^2}cdot frac{xcdot2^x-xcdot 3^x}{1+xcdot3^x}}}= ldots$$



                              since you take the limit for a part of the expression.



                              Refer also to the related




                              • Problem with limit solving

                              • Analyzing limits problem Calculus (tell me where I'm wrong).







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Nov 18 at 8:12

























                              answered Nov 18 at 7:59









                              gimusi

                              87.7k74393




                              87.7k74393












                              • 1. Don't arithmetical properties of limits allow us to take the limit for a part of the expression? 2. Would it had been correct if I have sent the "lim" into the exponent? Thank you!
                                – fragileradius
                                Nov 18 at 8:08






                              • 1




                                @fragileradius I've added some references with a full discussion about that.
                                – gimusi
                                Nov 18 at 8:12


















                              • 1. Don't arithmetical properties of limits allow us to take the limit for a part of the expression? 2. Would it had been correct if I have sent the "lim" into the exponent? Thank you!
                                – fragileradius
                                Nov 18 at 8:08






                              • 1




                                @fragileradius I've added some references with a full discussion about that.
                                – gimusi
                                Nov 18 at 8:12
















                              1. Don't arithmetical properties of limits allow us to take the limit for a part of the expression? 2. Would it had been correct if I have sent the "lim" into the exponent? Thank you!
                              – fragileradius
                              Nov 18 at 8:08




                              1. Don't arithmetical properties of limits allow us to take the limit for a part of the expression? 2. Would it had been correct if I have sent the "lim" into the exponent? Thank you!
                              – fragileradius
                              Nov 18 at 8:08




                              1




                              1




                              @fragileradius I've added some references with a full discussion about that.
                              – gimusi
                              Nov 18 at 8:12




                              @fragileradius I've added some references with a full discussion about that.
                              – gimusi
                              Nov 18 at 8:12


















                               

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