Compute $limlimits_{xto0} left(sin x + cos xright)^{1/x}$












0












$begingroup$


I hit a snag while solving exponential functions whose limits are given.



Question:



$$lim_{xto0} left(sin x + cos xright)^left(1/xright)$$



My Approach:



I am using the followin relation to solve the question of these type.



$$lim_{xto0} left(1 + xright)^left(1/xright) = e qquad(2)$$



But now how should i convert my above question so that i can apply the rule as mentioned in $(2)$.



Conclusion:



First of all help will be appreciated.



Second how to solve functions of such kind in a quick method.



Thanks,



P.S.(Feel free to edit my question if you find any errors or mistakes in my question)






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Crostul The LH step in this case would be absurd, as in every situation where one is after a limit $limlimits_{xto0}g(x)/x$ with $g$ differentiable at $0$ and $g(0)=0$.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10










  • $begingroup$
    @Did , please elaborate why would the LH step be absurd?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:24






  • 2




    $begingroup$
    Because using LH to find $limlimits_{xto0} g(x)/h(x)$ when $g(0)=h(0)=0$ requires to compute $g'(0)$ and $h'(0)$. But, in the case when $h(x)=x$, one already knows that $limlimits_{xto0}g(x)/x=g'(0)$ by definition of the derivative of $g$ since $g(x)/x=(g(x)-g(0))/(x-0)$. So, invoking LH here amounts to a beautifully (?) absurd logical circle.
    $endgroup$
    – Did
    Dec 31 '18 at 12:29


















0












$begingroup$


I hit a snag while solving exponential functions whose limits are given.



Question:



$$lim_{xto0} left(sin x + cos xright)^left(1/xright)$$



My Approach:



I am using the followin relation to solve the question of these type.



$$lim_{xto0} left(1 + xright)^left(1/xright) = e qquad(2)$$



But now how should i convert my above question so that i can apply the rule as mentioned in $(2)$.



Conclusion:



First of all help will be appreciated.



Second how to solve functions of such kind in a quick method.



Thanks,



P.S.(Feel free to edit my question if you find any errors or mistakes in my question)






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Crostul The LH step in this case would be absurd, as in every situation where one is after a limit $limlimits_{xto0}g(x)/x$ with $g$ differentiable at $0$ and $g(0)=0$.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10










  • $begingroup$
    @Did , please elaborate why would the LH step be absurd?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:24






  • 2




    $begingroup$
    Because using LH to find $limlimits_{xto0} g(x)/h(x)$ when $g(0)=h(0)=0$ requires to compute $g'(0)$ and $h'(0)$. But, in the case when $h(x)=x$, one already knows that $limlimits_{xto0}g(x)/x=g'(0)$ by definition of the derivative of $g$ since $g(x)/x=(g(x)-g(0))/(x-0)$. So, invoking LH here amounts to a beautifully (?) absurd logical circle.
    $endgroup$
    – Did
    Dec 31 '18 at 12:29
















0












0








0





$begingroup$


I hit a snag while solving exponential functions whose limits are given.



Question:



$$lim_{xto0} left(sin x + cos xright)^left(1/xright)$$



My Approach:



I am using the followin relation to solve the question of these type.



$$lim_{xto0} left(1 + xright)^left(1/xright) = e qquad(2)$$



But now how should i convert my above question so that i can apply the rule as mentioned in $(2)$.



Conclusion:



First of all help will be appreciated.



Second how to solve functions of such kind in a quick method.



Thanks,



P.S.(Feel free to edit my question if you find any errors or mistakes in my question)






limits







share|cite|improve this question











$endgroup$




I hit a snag while solving exponential functions whose limits are given.



Question:



$$lim_{xto0} left(sin x + cos xright)^left(1/xright)$$



My Approach:



I am using the followin relation to solve the question of these type.



$$lim_{xto0} left(1 + xright)^left(1/xright) = e qquad(2)$$



But now how should i convert my above question so that i can apply the rule as mentioned in $(2)$.



Conclusion:



First of all help will be appreciated.



Second how to solve functions of such kind in a quick method.



Thanks,



P.S.(Feel free to edit my question if you find any errors or mistakes in my question)






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 14:17









Fortox

657




657










asked Dec 31 '18 at 12:04









GingitsuneGingitsune

247




247












  • $begingroup$
    @Crostul The LH step in this case would be absurd, as in every situation where one is after a limit $limlimits_{xto0}g(x)/x$ with $g$ differentiable at $0$ and $g(0)=0$.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10










  • $begingroup$
    @Did , please elaborate why would the LH step be absurd?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:24






  • 2




    $begingroup$
    Because using LH to find $limlimits_{xto0} g(x)/h(x)$ when $g(0)=h(0)=0$ requires to compute $g'(0)$ and $h'(0)$. But, in the case when $h(x)=x$, one already knows that $limlimits_{xto0}g(x)/x=g'(0)$ by definition of the derivative of $g$ since $g(x)/x=(g(x)-g(0))/(x-0)$. So, invoking LH here amounts to a beautifully (?) absurd logical circle.
    $endgroup$
    – Did
    Dec 31 '18 at 12:29




















  • $begingroup$
    @Crostul The LH step in this case would be absurd, as in every situation where one is after a limit $limlimits_{xto0}g(x)/x$ with $g$ differentiable at $0$ and $g(0)=0$.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10










  • $begingroup$
    @Did , please elaborate why would the LH step be absurd?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:24






  • 2




    $begingroup$
    Because using LH to find $limlimits_{xto0} g(x)/h(x)$ when $g(0)=h(0)=0$ requires to compute $g'(0)$ and $h'(0)$. But, in the case when $h(x)=x$, one already knows that $limlimits_{xto0}g(x)/x=g'(0)$ by definition of the derivative of $g$ since $g(x)/x=(g(x)-g(0))/(x-0)$. So, invoking LH here amounts to a beautifully (?) absurd logical circle.
    $endgroup$
    – Did
    Dec 31 '18 at 12:29


















$begingroup$
@Crostul The LH step in this case would be absurd, as in every situation where one is after a limit $limlimits_{xto0}g(x)/x$ with $g$ differentiable at $0$ and $g(0)=0$.
$endgroup$
– Did
Dec 31 '18 at 12:10




$begingroup$
@Crostul The LH step in this case would be absurd, as in every situation where one is after a limit $limlimits_{xto0}g(x)/x$ with $g$ differentiable at $0$ and $g(0)=0$.
$endgroup$
– Did
Dec 31 '18 at 12:10












$begingroup$
@Did , please elaborate why would the LH step be absurd?
$endgroup$
– Gingitsune
Dec 31 '18 at 12:24




$begingroup$
@Did , please elaborate why would the LH step be absurd?
$endgroup$
– Gingitsune
Dec 31 '18 at 12:24




2




2




$begingroup$
Because using LH to find $limlimits_{xto0} g(x)/h(x)$ when $g(0)=h(0)=0$ requires to compute $g'(0)$ and $h'(0)$. But, in the case when $h(x)=x$, one already knows that $limlimits_{xto0}g(x)/x=g'(0)$ by definition of the derivative of $g$ since $g(x)/x=(g(x)-g(0))/(x-0)$. So, invoking LH here amounts to a beautifully (?) absurd logical circle.
$endgroup$
– Did
Dec 31 '18 at 12:29






$begingroup$
Because using LH to find $limlimits_{xto0} g(x)/h(x)$ when $g(0)=h(0)=0$ requires to compute $g'(0)$ and $h'(0)$. But, in the case when $h(x)=x$, one already knows that $limlimits_{xto0}g(x)/x=g'(0)$ by definition of the derivative of $g$ since $g(x)/x=(g(x)-g(0))/(x-0)$. So, invoking LH here amounts to a beautifully (?) absurd logical circle.
$endgroup$
– Did
Dec 31 '18 at 12:29












3 Answers
3






active

oldest

votes


















6












$begingroup$

In this case the best idea is take logarithm, and then use De l'Hopital. $$lim_{x to 0} frac{ln (sin x + cos x)}{x} = lim_{x to 0} frac{cos x - sin x}{sin x + cos x} = 1$$ Hence the answer is $e^{1}=e$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    See comment on main.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10






  • 1




    $begingroup$
    It isn't $e^{-1}$ anyway
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 12:19






  • 2




    $begingroup$
    @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:21



















2












$begingroup$

Alternatively, take $cos x$ out:
$$lim_{xto0} left(sin x + cos xright)^left(1/xright)=\
lim_{xto0} left(cos xright)^left(1/xright)cdot lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)}cdot color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)}=color{blue}1cdot color{red}e=e,$$
because using the relation you want:
$$color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)=\
lim_{xto0} left[left(1+left(-2sin^2 frac x2right)right)^frac{1}{-2sin^2 frac x2}right]^{frac{-2sin^2 frac x2}{x}}=e^0=1};\
color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
lim_{xto0} left[left(1+frac{sin x}{cos x}right)^{frac{cos x}{sin x}}right]^frac{sin x}{xcos x}=e^1=e}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks you helped me find one more way to approach questions of this kind
    $endgroup$
    – Gingitsune
    Jan 1 at 12:06



















0












$begingroup$

Another standard way works with enforcing a $color{red}{1}$ in the basis and considering the resulting exponent:



begin{eqnarray*} left(sin x + cos xright)^{1/x}
& = & left( color{red}{1} + (color{blue}{sin x + cos x - 1})right)^{1/x}\
& = & left (left( 1 + (color{blue}{sin x + cos x - 1})right)^{1/(color{blue}{sin x + cos x - 1})}right)^{frac{color{blue}{sin x + cos x - 1}}{x}} \
& stackrel {x to 0}{longrightarrow} & e^{lim_{xto 0}frac{sin x + cos x - 1}{x}}\
& = & e^{1+cos'(0)} = e \
end{eqnarray*}






share|cite|improve this answer









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    In this case the best idea is take logarithm, and then use De l'Hopital. $$lim_{x to 0} frac{ln (sin x + cos x)}{x} = lim_{x to 0} frac{cos x - sin x}{sin x + cos x} = 1$$ Hence the answer is $e^{1}=e$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      See comment on main.
      $endgroup$
      – Did
      Dec 31 '18 at 12:10






    • 1




      $begingroup$
      It isn't $e^{-1}$ anyway
      $endgroup$
      – Kenny Lau
      Dec 31 '18 at 12:19






    • 2




      $begingroup$
      @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
      $endgroup$
      – Gingitsune
      Dec 31 '18 at 12:21
















    6












    $begingroup$

    In this case the best idea is take logarithm, and then use De l'Hopital. $$lim_{x to 0} frac{ln (sin x + cos x)}{x} = lim_{x to 0} frac{cos x - sin x}{sin x + cos x} = 1$$ Hence the answer is $e^{1}=e$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      See comment on main.
      $endgroup$
      – Did
      Dec 31 '18 at 12:10






    • 1




      $begingroup$
      It isn't $e^{-1}$ anyway
      $endgroup$
      – Kenny Lau
      Dec 31 '18 at 12:19






    • 2




      $begingroup$
      @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
      $endgroup$
      – Gingitsune
      Dec 31 '18 at 12:21














    6












    6








    6





    $begingroup$

    In this case the best idea is take logarithm, and then use De l'Hopital. $$lim_{x to 0} frac{ln (sin x + cos x)}{x} = lim_{x to 0} frac{cos x - sin x}{sin x + cos x} = 1$$ Hence the answer is $e^{1}=e$.






    share|cite|improve this answer











    $endgroup$



    In this case the best idea is take logarithm, and then use De l'Hopital. $$lim_{x to 0} frac{ln (sin x + cos x)}{x} = lim_{x to 0} frac{cos x - sin x}{sin x + cos x} = 1$$ Hence the answer is $e^{1}=e$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 31 '18 at 12:40

























    answered Dec 31 '18 at 12:08









    CrostulCrostul

    27.8k22352




    27.8k22352












    • $begingroup$
      See comment on main.
      $endgroup$
      – Did
      Dec 31 '18 at 12:10






    • 1




      $begingroup$
      It isn't $e^{-1}$ anyway
      $endgroup$
      – Kenny Lau
      Dec 31 '18 at 12:19






    • 2




      $begingroup$
      @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
      $endgroup$
      – Gingitsune
      Dec 31 '18 at 12:21


















    • $begingroup$
      See comment on main.
      $endgroup$
      – Did
      Dec 31 '18 at 12:10






    • 1




      $begingroup$
      It isn't $e^{-1}$ anyway
      $endgroup$
      – Kenny Lau
      Dec 31 '18 at 12:19






    • 2




      $begingroup$
      @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
      $endgroup$
      – Gingitsune
      Dec 31 '18 at 12:21
















    $begingroup$
    See comment on main.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10




    $begingroup$
    See comment on main.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10




    1




    1




    $begingroup$
    It isn't $e^{-1}$ anyway
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 12:19




    $begingroup$
    It isn't $e^{-1}$ anyway
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 12:19




    2




    2




    $begingroup$
    @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:21




    $begingroup$
    @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:21











    2












    $begingroup$

    Alternatively, take $cos x$ out:
    $$lim_{xto0} left(sin x + cos xright)^left(1/xright)=\
    lim_{xto0} left(cos xright)^left(1/xright)cdot lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)}cdot color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)}=color{blue}1cdot color{red}e=e,$$
    because using the relation you want:
    $$color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)=\
    lim_{xto0} left[left(1+left(-2sin^2 frac x2right)right)^frac{1}{-2sin^2 frac x2}right]^{frac{-2sin^2 frac x2}{x}}=e^0=1};\
    color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    lim_{xto0} left[left(1+frac{sin x}{cos x}right)^{frac{cos x}{sin x}}right]^frac{sin x}{xcos x}=e^1=e}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thanks you helped me find one more way to approach questions of this kind
      $endgroup$
      – Gingitsune
      Jan 1 at 12:06
















    2












    $begingroup$

    Alternatively, take $cos x$ out:
    $$lim_{xto0} left(sin x + cos xright)^left(1/xright)=\
    lim_{xto0} left(cos xright)^left(1/xright)cdot lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)}cdot color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)}=color{blue}1cdot color{red}e=e,$$
    because using the relation you want:
    $$color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)=\
    lim_{xto0} left[left(1+left(-2sin^2 frac x2right)right)^frac{1}{-2sin^2 frac x2}right]^{frac{-2sin^2 frac x2}{x}}=e^0=1};\
    color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    lim_{xto0} left[left(1+frac{sin x}{cos x}right)^{frac{cos x}{sin x}}right]^frac{sin x}{xcos x}=e^1=e}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thanks you helped me find one more way to approach questions of this kind
      $endgroup$
      – Gingitsune
      Jan 1 at 12:06














    2












    2








    2





    $begingroup$

    Alternatively, take $cos x$ out:
    $$lim_{xto0} left(sin x + cos xright)^left(1/xright)=\
    lim_{xto0} left(cos xright)^left(1/xright)cdot lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)}cdot color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)}=color{blue}1cdot color{red}e=e,$$
    because using the relation you want:
    $$color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)=\
    lim_{xto0} left[left(1+left(-2sin^2 frac x2right)right)^frac{1}{-2sin^2 frac x2}right]^{frac{-2sin^2 frac x2}{x}}=e^0=1};\
    color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    lim_{xto0} left[left(1+frac{sin x}{cos x}right)^{frac{cos x}{sin x}}right]^frac{sin x}{xcos x}=e^1=e}.$$






    share|cite|improve this answer









    $endgroup$



    Alternatively, take $cos x$ out:
    $$lim_{xto0} left(sin x + cos xright)^left(1/xright)=\
    lim_{xto0} left(cos xright)^left(1/xright)cdot lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)}cdot color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)}=color{blue}1cdot color{red}e=e,$$
    because using the relation you want:
    $$color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)=\
    lim_{xto0} left[left(1+left(-2sin^2 frac x2right)right)^frac{1}{-2sin^2 frac x2}right]^{frac{-2sin^2 frac x2}{x}}=e^0=1};\
    color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    lim_{xto0} left[left(1+frac{sin x}{cos x}right)^{frac{cos x}{sin x}}right]^frac{sin x}{xcos x}=e^1=e}.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 1 at 10:05









    farruhotafarruhota

    19.7k2738




    19.7k2738












    • $begingroup$
      thanks you helped me find one more way to approach questions of this kind
      $endgroup$
      – Gingitsune
      Jan 1 at 12:06


















    • $begingroup$
      thanks you helped me find one more way to approach questions of this kind
      $endgroup$
      – Gingitsune
      Jan 1 at 12:06
















    $begingroup$
    thanks you helped me find one more way to approach questions of this kind
    $endgroup$
    – Gingitsune
    Jan 1 at 12:06




    $begingroup$
    thanks you helped me find one more way to approach questions of this kind
    $endgroup$
    – Gingitsune
    Jan 1 at 12:06











    0












    $begingroup$

    Another standard way works with enforcing a $color{red}{1}$ in the basis and considering the resulting exponent:



    begin{eqnarray*} left(sin x + cos xright)^{1/x}
    & = & left( color{red}{1} + (color{blue}{sin x + cos x - 1})right)^{1/x}\
    & = & left (left( 1 + (color{blue}{sin x + cos x - 1})right)^{1/(color{blue}{sin x + cos x - 1})}right)^{frac{color{blue}{sin x + cos x - 1}}{x}} \
    & stackrel {x to 0}{longrightarrow} & e^{lim_{xto 0}frac{sin x + cos x - 1}{x}}\
    & = & e^{1+cos'(0)} = e \
    end{eqnarray*}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Another standard way works with enforcing a $color{red}{1}$ in the basis and considering the resulting exponent:



      begin{eqnarray*} left(sin x + cos xright)^{1/x}
      & = & left( color{red}{1} + (color{blue}{sin x + cos x - 1})right)^{1/x}\
      & = & left (left( 1 + (color{blue}{sin x + cos x - 1})right)^{1/(color{blue}{sin x + cos x - 1})}right)^{frac{color{blue}{sin x + cos x - 1}}{x}} \
      & stackrel {x to 0}{longrightarrow} & e^{lim_{xto 0}frac{sin x + cos x - 1}{x}}\
      & = & e^{1+cos'(0)} = e \
      end{eqnarray*}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Another standard way works with enforcing a $color{red}{1}$ in the basis and considering the resulting exponent:



        begin{eqnarray*} left(sin x + cos xright)^{1/x}
        & = & left( color{red}{1} + (color{blue}{sin x + cos x - 1})right)^{1/x}\
        & = & left (left( 1 + (color{blue}{sin x + cos x - 1})right)^{1/(color{blue}{sin x + cos x - 1})}right)^{frac{color{blue}{sin x + cos x - 1}}{x}} \
        & stackrel {x to 0}{longrightarrow} & e^{lim_{xto 0}frac{sin x + cos x - 1}{x}}\
        & = & e^{1+cos'(0)} = e \
        end{eqnarray*}






        share|cite|improve this answer









        $endgroup$



        Another standard way works with enforcing a $color{red}{1}$ in the basis and considering the resulting exponent:



        begin{eqnarray*} left(sin x + cos xright)^{1/x}
        & = & left( color{red}{1} + (color{blue}{sin x + cos x - 1})right)^{1/x}\
        & = & left (left( 1 + (color{blue}{sin x + cos x - 1})right)^{1/(color{blue}{sin x + cos x - 1})}right)^{frac{color{blue}{sin x + cos x - 1}}{x}} \
        & stackrel {x to 0}{longrightarrow} & e^{lim_{xto 0}frac{sin x + cos x - 1}{x}}\
        & = & e^{1+cos'(0)} = e \
        end{eqnarray*}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 6:11









        trancelocationtrancelocation

        9,8101722




        9,8101722






























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