The idea of having a limit as $x rightarrow a$ where the denominator is 0 confuses me












0












$begingroup$


What does that mean? Say as $x rightarrow 2$, our equation looks like this, $2x + 1 over {x - 2}$.



Then the denominator will go towards 0. Now how would I find the limit of this? How do I know it will exist?



Also, can someone help me know why the limit of $e^x over {e^x + 1}$ as $x rightarrow -infty$ is $0$? and when it's approaching $+ infty$ the limit is 1? And why that represents the horizontal asymptote?






calculus







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$endgroup$

















    0












    $begingroup$


    What does that mean? Say as $x rightarrow 2$, our equation looks like this, $2x + 1 over {x - 2}$.



    Then the denominator will go towards 0. Now how would I find the limit of this? How do I know it will exist?



    Also, can someone help me know why the limit of $e^x over {e^x + 1}$ as $x rightarrow -infty$ is $0$? and when it's approaching $+ infty$ the limit is 1? And why that represents the horizontal asymptote?






    calculus







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      What does that mean? Say as $x rightarrow 2$, our equation looks like this, $2x + 1 over {x - 2}$.



      Then the denominator will go towards 0. Now how would I find the limit of this? How do I know it will exist?



      Also, can someone help me know why the limit of $e^x over {e^x + 1}$ as $x rightarrow -infty$ is $0$? and when it's approaching $+ infty$ the limit is 1? And why that represents the horizontal asymptote?






      calculus







      share|cite|improve this question









      $endgroup$




      What does that mean? Say as $x rightarrow 2$, our equation looks like this, $2x + 1 over {x - 2}$.



      Then the denominator will go towards 0. Now how would I find the limit of this? How do I know it will exist?



      Also, can someone help me know why the limit of $e^x over {e^x + 1}$ as $x rightarrow -infty$ is $0$? and when it's approaching $+ infty$ the limit is 1? And why that represents the horizontal asymptote?






      calculus




      calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 5:53









      mingming

      3956




      3956






















          2 Answers
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          1












          $begingroup$

          First off, $dfrac{2x+1}{x-2}$ is not an "equation" since there is no equal sign. You probably want to call that a mathematical expression or a function (and then you "take the limit" of the function as $xrightarrow 2$).



          In order for the "bilateral" limit as $xrightarrow 2$ to exist, the "unilateral" limits (left and right) must exist and be equal.



          Replacing $x$ by $2^-$ yields $2^--2=0^-$ in which case your function takes the form $dfrac{text{+}}{0^-}rightarrow-infty$



          Replacing $x$ by $2^+$ yields $2^+-2=0^+$ in which case your function takes the form $dfrac{text{+}}{0^+}rightarrow+infty$



          The unilateral limits are not equal (I like to say they "don't agree") so the limit as $xrightarrow2$ does not exist.



          Another suggestion for the second part:



          $$f(x)=frac{e^x}{e^x+1}=frac{e^x}{e^x(1+e^{-x})}=frac{1}{1+e^{-x}}$$



          If $xrightarrow+infty$, then $f(x)rightarrowfrac{1}{1+e^{-infty}}=frac{1}{1+0}=1$



          If $xrightarrow-infty$, then $f(x)rightarrowfrac{1}{1+e^{-(-infty))}}=frac{1}{1+infty}=0$



          To clarify, if you have $frac{0}{0}$ or $pmfrac{infty}{infty}$, it's indeterminate and factoring might help. Otherwise:



          $$frac{+}{0^+}rightarrow+infty,quadfrac{-}{0^+}rightarrow-infty,quadfrac{+}{0^-}rightarrow-infty,quadfrac{-}{0^-}rightarrow+infty,$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
            $endgroup$
            – ming
            Dec 10 '18 at 6:31










          • $begingroup$
            Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
            $endgroup$
            – ming
            Dec 10 '18 at 6:34










          • $begingroup$
            @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
            $endgroup$
            – orion2112
            Dec 10 '18 at 6:36










          • $begingroup$
            @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
            $endgroup$
            – orion2112
            Dec 10 '18 at 6:45










          • $begingroup$
            Cool! So only use both sided limits if we see division by 0 or a piecewise function.
            $endgroup$
            – ming
            Dec 10 '18 at 6:48



















          0












          $begingroup$

          For $frac{2x+1}{x-2}$, it may help to sketch a plot by yourself. When $x$ approaches $2$ from above, $frac{2x+1}{x-2}$ gets arbitrarily large.
          When $x$ approaches $2$ from below, $frac{2x+1}{x-2}$ gets arbitrarily small (negative). So, the limit does not exist.




          For $frac{e^x}{e^x+1}$, if $x to -infty$, then $e^x to 0$, so
          $frac{e^x}{e^x+1} to frac{0}{0+1}$.



          When taking $x to infty$, it may help to write $frac{e^x}{e^x+1} = 1 - frac{1}{e^x+1}$.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

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            active

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            active

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            1












            $begingroup$

            First off, $dfrac{2x+1}{x-2}$ is not an "equation" since there is no equal sign. You probably want to call that a mathematical expression or a function (and then you "take the limit" of the function as $xrightarrow 2$).



            In order for the "bilateral" limit as $xrightarrow 2$ to exist, the "unilateral" limits (left and right) must exist and be equal.



            Replacing $x$ by $2^-$ yields $2^--2=0^-$ in which case your function takes the form $dfrac{text{+}}{0^-}rightarrow-infty$



            Replacing $x$ by $2^+$ yields $2^+-2=0^+$ in which case your function takes the form $dfrac{text{+}}{0^+}rightarrow+infty$



            The unilateral limits are not equal (I like to say they "don't agree") so the limit as $xrightarrow2$ does not exist.



            Another suggestion for the second part:



            $$f(x)=frac{e^x}{e^x+1}=frac{e^x}{e^x(1+e^{-x})}=frac{1}{1+e^{-x}}$$



            If $xrightarrow+infty$, then $f(x)rightarrowfrac{1}{1+e^{-infty}}=frac{1}{1+0}=1$



            If $xrightarrow-infty$, then $f(x)rightarrowfrac{1}{1+e^{-(-infty))}}=frac{1}{1+infty}=0$



            To clarify, if you have $frac{0}{0}$ or $pmfrac{infty}{infty}$, it's indeterminate and factoring might help. Otherwise:



            $$frac{+}{0^+}rightarrow+infty,quadfrac{-}{0^+}rightarrow-infty,quadfrac{+}{0^-}rightarrow-infty,quadfrac{-}{0^-}rightarrow+infty,$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
              $endgroup$
              – ming
              Dec 10 '18 at 6:31










            • $begingroup$
              Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
              $endgroup$
              – ming
              Dec 10 '18 at 6:34










            • $begingroup$
              @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:36










            • $begingroup$
              @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:45










            • $begingroup$
              Cool! So only use both sided limits if we see division by 0 or a piecewise function.
              $endgroup$
              – ming
              Dec 10 '18 at 6:48
















            1












            $begingroup$

            First off, $dfrac{2x+1}{x-2}$ is not an "equation" since there is no equal sign. You probably want to call that a mathematical expression or a function (and then you "take the limit" of the function as $xrightarrow 2$).



            In order for the "bilateral" limit as $xrightarrow 2$ to exist, the "unilateral" limits (left and right) must exist and be equal.



            Replacing $x$ by $2^-$ yields $2^--2=0^-$ in which case your function takes the form $dfrac{text{+}}{0^-}rightarrow-infty$



            Replacing $x$ by $2^+$ yields $2^+-2=0^+$ in which case your function takes the form $dfrac{text{+}}{0^+}rightarrow+infty$



            The unilateral limits are not equal (I like to say they "don't agree") so the limit as $xrightarrow2$ does not exist.



            Another suggestion for the second part:



            $$f(x)=frac{e^x}{e^x+1}=frac{e^x}{e^x(1+e^{-x})}=frac{1}{1+e^{-x}}$$



            If $xrightarrow+infty$, then $f(x)rightarrowfrac{1}{1+e^{-infty}}=frac{1}{1+0}=1$



            If $xrightarrow-infty$, then $f(x)rightarrowfrac{1}{1+e^{-(-infty))}}=frac{1}{1+infty}=0$



            To clarify, if you have $frac{0}{0}$ or $pmfrac{infty}{infty}$, it's indeterminate and factoring might help. Otherwise:



            $$frac{+}{0^+}rightarrow+infty,quadfrac{-}{0^+}rightarrow-infty,quadfrac{+}{0^-}rightarrow-infty,quadfrac{-}{0^-}rightarrow+infty,$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
              $endgroup$
              – ming
              Dec 10 '18 at 6:31










            • $begingroup$
              Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
              $endgroup$
              – ming
              Dec 10 '18 at 6:34










            • $begingroup$
              @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:36










            • $begingroup$
              @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:45










            • $begingroup$
              Cool! So only use both sided limits if we see division by 0 or a piecewise function.
              $endgroup$
              – ming
              Dec 10 '18 at 6:48














            1












            1








            1





            $begingroup$

            First off, $dfrac{2x+1}{x-2}$ is not an "equation" since there is no equal sign. You probably want to call that a mathematical expression or a function (and then you "take the limit" of the function as $xrightarrow 2$).



            In order for the "bilateral" limit as $xrightarrow 2$ to exist, the "unilateral" limits (left and right) must exist and be equal.



            Replacing $x$ by $2^-$ yields $2^--2=0^-$ in which case your function takes the form $dfrac{text{+}}{0^-}rightarrow-infty$



            Replacing $x$ by $2^+$ yields $2^+-2=0^+$ in which case your function takes the form $dfrac{text{+}}{0^+}rightarrow+infty$



            The unilateral limits are not equal (I like to say they "don't agree") so the limit as $xrightarrow2$ does not exist.



            Another suggestion for the second part:



            $$f(x)=frac{e^x}{e^x+1}=frac{e^x}{e^x(1+e^{-x})}=frac{1}{1+e^{-x}}$$



            If $xrightarrow+infty$, then $f(x)rightarrowfrac{1}{1+e^{-infty}}=frac{1}{1+0}=1$



            If $xrightarrow-infty$, then $f(x)rightarrowfrac{1}{1+e^{-(-infty))}}=frac{1}{1+infty}=0$



            To clarify, if you have $frac{0}{0}$ or $pmfrac{infty}{infty}$, it's indeterminate and factoring might help. Otherwise:



            $$frac{+}{0^+}rightarrow+infty,quadfrac{-}{0^+}rightarrow-infty,quadfrac{+}{0^-}rightarrow-infty,quadfrac{-}{0^-}rightarrow+infty,$$






            share|cite|improve this answer











            $endgroup$



            First off, $dfrac{2x+1}{x-2}$ is not an "equation" since there is no equal sign. You probably want to call that a mathematical expression or a function (and then you "take the limit" of the function as $xrightarrow 2$).



            In order for the "bilateral" limit as $xrightarrow 2$ to exist, the "unilateral" limits (left and right) must exist and be equal.



            Replacing $x$ by $2^-$ yields $2^--2=0^-$ in which case your function takes the form $dfrac{text{+}}{0^-}rightarrow-infty$



            Replacing $x$ by $2^+$ yields $2^+-2=0^+$ in which case your function takes the form $dfrac{text{+}}{0^+}rightarrow+infty$



            The unilateral limits are not equal (I like to say they "don't agree") so the limit as $xrightarrow2$ does not exist.



            Another suggestion for the second part:



            $$f(x)=frac{e^x}{e^x+1}=frac{e^x}{e^x(1+e^{-x})}=frac{1}{1+e^{-x}}$$



            If $xrightarrow+infty$, then $f(x)rightarrowfrac{1}{1+e^{-infty}}=frac{1}{1+0}=1$



            If $xrightarrow-infty$, then $f(x)rightarrowfrac{1}{1+e^{-(-infty))}}=frac{1}{1+infty}=0$



            To clarify, if you have $frac{0}{0}$ or $pmfrac{infty}{infty}$, it's indeterminate and factoring might help. Otherwise:



            $$frac{+}{0^+}rightarrow+infty,quadfrac{-}{0^+}rightarrow-infty,quadfrac{+}{0^-}rightarrow-infty,quadfrac{-}{0^-}rightarrow+infty,$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 10 '18 at 7:06

























            answered Dec 10 '18 at 6:21









            orion2112orion2112

            471310




            471310












            • $begingroup$
              So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
              $endgroup$
              – ming
              Dec 10 '18 at 6:31










            • $begingroup$
              Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
              $endgroup$
              – ming
              Dec 10 '18 at 6:34










            • $begingroup$
              @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:36










            • $begingroup$
              @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:45










            • $begingroup$
              Cool! So only use both sided limits if we see division by 0 or a piecewise function.
              $endgroup$
              – ming
              Dec 10 '18 at 6:48


















            • $begingroup$
              So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
              $endgroup$
              – ming
              Dec 10 '18 at 6:31










            • $begingroup$
              Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
              $endgroup$
              – ming
              Dec 10 '18 at 6:34










            • $begingroup$
              @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:36










            • $begingroup$
              @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
              $endgroup$
              – orion2112
              Dec 10 '18 at 6:45










            • $begingroup$
              Cool! So only use both sided limits if we see division by 0 or a piecewise function.
              $endgroup$
              – ming
              Dec 10 '18 at 6:48
















            $begingroup$
            So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
            $endgroup$
            – ming
            Dec 10 '18 at 6:31




            $begingroup$
            So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions?
            $endgroup$
            – ming
            Dec 10 '18 at 6:31












            $begingroup$
            Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
            $endgroup$
            – ming
            Dec 10 '18 at 6:34




            $begingroup$
            Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule?
            $endgroup$
            – ming
            Dec 10 '18 at 6:34












            $begingroup$
            @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
            $endgroup$
            – orion2112
            Dec 10 '18 at 6:36




            $begingroup$
            @ming More specifically, these last 2 limits indicate the behavior of the function near $+infty$ and $-infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$.
            $endgroup$
            – orion2112
            Dec 10 '18 at 6:36












            $begingroup$
            @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
            $endgroup$
            – orion2112
            Dec 10 '18 at 6:45




            $begingroup$
            @ming You can't use the left/right limit when $xrightarrowpminfty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts.
            $endgroup$
            – orion2112
            Dec 10 '18 at 6:45












            $begingroup$
            Cool! So only use both sided limits if we see division by 0 or a piecewise function.
            $endgroup$
            – ming
            Dec 10 '18 at 6:48




            $begingroup$
            Cool! So only use both sided limits if we see division by 0 or a piecewise function.
            $endgroup$
            – ming
            Dec 10 '18 at 6:48











            0












            $begingroup$

            For $frac{2x+1}{x-2}$, it may help to sketch a plot by yourself. When $x$ approaches $2$ from above, $frac{2x+1}{x-2}$ gets arbitrarily large.
            When $x$ approaches $2$ from below, $frac{2x+1}{x-2}$ gets arbitrarily small (negative). So, the limit does not exist.




            For $frac{e^x}{e^x+1}$, if $x to -infty$, then $e^x to 0$, so
            $frac{e^x}{e^x+1} to frac{0}{0+1}$.



            When taking $x to infty$, it may help to write $frac{e^x}{e^x+1} = 1 - frac{1}{e^x+1}$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              For $frac{2x+1}{x-2}$, it may help to sketch a plot by yourself. When $x$ approaches $2$ from above, $frac{2x+1}{x-2}$ gets arbitrarily large.
              When $x$ approaches $2$ from below, $frac{2x+1}{x-2}$ gets arbitrarily small (negative). So, the limit does not exist.




              For $frac{e^x}{e^x+1}$, if $x to -infty$, then $e^x to 0$, so
              $frac{e^x}{e^x+1} to frac{0}{0+1}$.



              When taking $x to infty$, it may help to write $frac{e^x}{e^x+1} = 1 - frac{1}{e^x+1}$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                For $frac{2x+1}{x-2}$, it may help to sketch a plot by yourself. When $x$ approaches $2$ from above, $frac{2x+1}{x-2}$ gets arbitrarily large.
                When $x$ approaches $2$ from below, $frac{2x+1}{x-2}$ gets arbitrarily small (negative). So, the limit does not exist.




                For $frac{e^x}{e^x+1}$, if $x to -infty$, then $e^x to 0$, so
                $frac{e^x}{e^x+1} to frac{0}{0+1}$.



                When taking $x to infty$, it may help to write $frac{e^x}{e^x+1} = 1 - frac{1}{e^x+1}$.






                share|cite|improve this answer









                $endgroup$



                For $frac{2x+1}{x-2}$, it may help to sketch a plot by yourself. When $x$ approaches $2$ from above, $frac{2x+1}{x-2}$ gets arbitrarily large.
                When $x$ approaches $2$ from below, $frac{2x+1}{x-2}$ gets arbitrarily small (negative). So, the limit does not exist.




                For $frac{e^x}{e^x+1}$, if $x to -infty$, then $e^x to 0$, so
                $frac{e^x}{e^x+1} to frac{0}{0+1}$.



                When taking $x to infty$, it may help to write $frac{e^x}{e^x+1} = 1 - frac{1}{e^x+1}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 10 '18 at 6:01









                angryavianangryavian

                42.3k23481




                42.3k23481






























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