Solving a basic limit - do not know the solution












0












$begingroup$


So the limit is:



$lim_{x rightarrow 2} frac{2x-4}{x^2+2x-6}$



So is it zero from the beginning. Because 0/something is zero, but that is stinky. And I am practicing from the book I do not have solution.
Symbolab an Wolframalpha are also saying it is zero.



But if it is taken with L'Hospital's rule it is 1/3. I do not get what am I doing wrong. I want to solve it adjusting the polynomials.



Thanks.






limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    the denominator is not zero,so you can't apply L'hospital rule
    $endgroup$
    – haqnatural
    Jan 24 at 20:40










  • $begingroup$
    You can only apply l’Hospital’s rule in certain situations; this is not one of them. $0$ is correct.
    $endgroup$
    – Clayton
    Jan 24 at 20:40










  • $begingroup$
    "0/something is zero": provided that something is not also zero.
    $endgroup$
    – Yves Daoust
    Jan 24 at 21:08
















0












$begingroup$


So the limit is:



$lim_{x rightarrow 2} frac{2x-4}{x^2+2x-6}$



So is it zero from the beginning. Because 0/something is zero, but that is stinky. And I am practicing from the book I do not have solution.
Symbolab an Wolframalpha are also saying it is zero.



But if it is taken with L'Hospital's rule it is 1/3. I do not get what am I doing wrong. I want to solve it adjusting the polynomials.



Thanks.






limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    the denominator is not zero,so you can't apply L'hospital rule
    $endgroup$
    – haqnatural
    Jan 24 at 20:40










  • $begingroup$
    You can only apply l’Hospital’s rule in certain situations; this is not one of them. $0$ is correct.
    $endgroup$
    – Clayton
    Jan 24 at 20:40










  • $begingroup$
    "0/something is zero": provided that something is not also zero.
    $endgroup$
    – Yves Daoust
    Jan 24 at 21:08














0












0








0





$begingroup$


So the limit is:



$lim_{x rightarrow 2} frac{2x-4}{x^2+2x-6}$



So is it zero from the beginning. Because 0/something is zero, but that is stinky. And I am practicing from the book I do not have solution.
Symbolab an Wolframalpha are also saying it is zero.



But if it is taken with L'Hospital's rule it is 1/3. I do not get what am I doing wrong. I want to solve it adjusting the polynomials.



Thanks.






limits







share|cite|improve this question











$endgroup$




So the limit is:



$lim_{x rightarrow 2} frac{2x-4}{x^2+2x-6}$



So is it zero from the beginning. Because 0/something is zero, but that is stinky. And I am practicing from the book I do not have solution.
Symbolab an Wolframalpha are also saying it is zero.



But if it is taken with L'Hospital's rule it is 1/3. I do not get what am I doing wrong. I want to solve it adjusting the polynomials.



Thanks.






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 20:44









Bernard

123k741116




123k741116










asked Jan 24 at 20:38









solidbastardsolidbastard

1




1








  • 1




    $begingroup$
    the denominator is not zero,so you can't apply L'hospital rule
    $endgroup$
    – haqnatural
    Jan 24 at 20:40










  • $begingroup$
    You can only apply l’Hospital’s rule in certain situations; this is not one of them. $0$ is correct.
    $endgroup$
    – Clayton
    Jan 24 at 20:40










  • $begingroup$
    "0/something is zero": provided that something is not also zero.
    $endgroup$
    – Yves Daoust
    Jan 24 at 21:08














  • 1




    $begingroup$
    the denominator is not zero,so you can't apply L'hospital rule
    $endgroup$
    – haqnatural
    Jan 24 at 20:40










  • $begingroup$
    You can only apply l’Hospital’s rule in certain situations; this is not one of them. $0$ is correct.
    $endgroup$
    – Clayton
    Jan 24 at 20:40










  • $begingroup$
    "0/something is zero": provided that something is not also zero.
    $endgroup$
    – Yves Daoust
    Jan 24 at 21:08








1




1




$begingroup$
the denominator is not zero,so you can't apply L'hospital rule
$endgroup$
– haqnatural
Jan 24 at 20:40




$begingroup$
the denominator is not zero,so you can't apply L'hospital rule
$endgroup$
– haqnatural
Jan 24 at 20:40












$begingroup$
You can only apply l’Hospital’s rule in certain situations; this is not one of them. $0$ is correct.
$endgroup$
– Clayton
Jan 24 at 20:40




$begingroup$
You can only apply l’Hospital’s rule in certain situations; this is not one of them. $0$ is correct.
$endgroup$
– Clayton
Jan 24 at 20:40












$begingroup$
"0/something is zero": provided that something is not also zero.
$endgroup$
– Yves Daoust
Jan 24 at 21:08




$begingroup$
"0/something is zero": provided that something is not also zero.
$endgroup$
– Yves Daoust
Jan 24 at 21:08










4 Answers
4






active

oldest

votes


















2












$begingroup$

You can't use L'Hospital's rule as there does not exist $frac{0}{0}$ or other specific condition for L'Hospital's rule. Hence $0$ is the correct value for the limit.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Since the denominator is non-zero, then you can't use L'Hopital's rule.
    As a first plan, when you have a rational function like this, plug x=2 in the numerator and denominator, if the denominator isn't 0, then you are good to go. If not, you may have to use L'Hopital (provided you can), or any other tool at your disposal.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      If $f(a)=0$ while $g(a)ne0$ then



      $$lim_{xto a}frac{f(x)}{g(x)}=0$$ because in the vicinity of $a$, the ratio get smaller and smaller.



      If $f(a)=0$ while $g(a)=0$, you cannot say (because the ratio could take any value). Anyway in this case L'Hospital pertains
      $$lim_{xto a}frac{f(x)}{g(x)}=lim_{xto a}frac{f'(x)}{g'(x)}.$$



      Compare



      enter image description here



      and



      enter image description here






      share|cite|improve this answer











      $endgroup$





















        0












        $begingroup$


        What are Limits?



        Limits are the way to find the value of a function when we get an
        indeterminate form of a number (like $frac 0 0$ which cannot be
        represented in Real number Line Axis).



        When we plot such functions where there is an indeterminate form at a
        certain value of $x$, then we don't get holes or gaps (try yourself!).
        Why? because they are not defined at a point but everywhere near them,
        so, that point is indefinitely small.



        Limits are the mathematical way to say almost when it's too small, so,




        unless you get a number in the indeterminate form, limits are not useful in such cases. So, next time, just simply put the value and evaluate the function like just any other function ;)



        Here in your case when we plug the value of $x$, we simply get 0, that's what the limit of the function is!



        Note: It's also possible that Limit of a function is not defined at a point, but the direct value exists! think yourself for such examples :P :)






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          You can't use L'Hospital's rule as there does not exist $frac{0}{0}$ or other specific condition for L'Hospital's rule. Hence $0$ is the correct value for the limit.






          share|cite|improve this answer











          $endgroup$


















            2












            $begingroup$

            You can't use L'Hospital's rule as there does not exist $frac{0}{0}$ or other specific condition for L'Hospital's rule. Hence $0$ is the correct value for the limit.






            share|cite|improve this answer











            $endgroup$
















              2












              2








              2





              $begingroup$

              You can't use L'Hospital's rule as there does not exist $frac{0}{0}$ or other specific condition for L'Hospital's rule. Hence $0$ is the correct value for the limit.






              share|cite|improve this answer











              $endgroup$



              You can't use L'Hospital's rule as there does not exist $frac{0}{0}$ or other specific condition for L'Hospital's rule. Hence $0$ is the correct value for the limit.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 24 at 20:43









              Bernard

              123k741116




              123k741116










              answered Jan 24 at 20:40









              OmGOmG

              2,512824




              2,512824























                  1












                  $begingroup$

                  Since the denominator is non-zero, then you can't use L'Hopital's rule.
                  As a first plan, when you have a rational function like this, plug x=2 in the numerator and denominator, if the denominator isn't 0, then you are good to go. If not, you may have to use L'Hopital (provided you can), or any other tool at your disposal.






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Since the denominator is non-zero, then you can't use L'Hopital's rule.
                    As a first plan, when you have a rational function like this, plug x=2 in the numerator and denominator, if the denominator isn't 0, then you are good to go. If not, you may have to use L'Hopital (provided you can), or any other tool at your disposal.






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Since the denominator is non-zero, then you can't use L'Hopital's rule.
                      As a first plan, when you have a rational function like this, plug x=2 in the numerator and denominator, if the denominator isn't 0, then you are good to go. If not, you may have to use L'Hopital (provided you can), or any other tool at your disposal.






                      share|cite|improve this answer









                      $endgroup$



                      Since the denominator is non-zero, then you can't use L'Hopital's rule.
                      As a first plan, when you have a rational function like this, plug x=2 in the numerator and denominator, if the denominator isn't 0, then you are good to go. If not, you may have to use L'Hopital (provided you can), or any other tool at your disposal.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 24 at 20:43









                      AlexandrosAlexandros

                      9421412




                      9421412























                          0












                          $begingroup$

                          If $f(a)=0$ while $g(a)ne0$ then



                          $$lim_{xto a}frac{f(x)}{g(x)}=0$$ because in the vicinity of $a$, the ratio get smaller and smaller.



                          If $f(a)=0$ while $g(a)=0$, you cannot say (because the ratio could take any value). Anyway in this case L'Hospital pertains
                          $$lim_{xto a}frac{f(x)}{g(x)}=lim_{xto a}frac{f'(x)}{g'(x)}.$$



                          Compare



                          enter image description here



                          and



                          enter image description here






                          share|cite|improve this answer











                          $endgroup$


















                            0












                            $begingroup$

                            If $f(a)=0$ while $g(a)ne0$ then



                            $$lim_{xto a}frac{f(x)}{g(x)}=0$$ because in the vicinity of $a$, the ratio get smaller and smaller.



                            If $f(a)=0$ while $g(a)=0$, you cannot say (because the ratio could take any value). Anyway in this case L'Hospital pertains
                            $$lim_{xto a}frac{f(x)}{g(x)}=lim_{xto a}frac{f'(x)}{g'(x)}.$$



                            Compare



                            enter image description here



                            and



                            enter image description here






                            share|cite|improve this answer











                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              If $f(a)=0$ while $g(a)ne0$ then



                              $$lim_{xto a}frac{f(x)}{g(x)}=0$$ because in the vicinity of $a$, the ratio get smaller and smaller.



                              If $f(a)=0$ while $g(a)=0$, you cannot say (because the ratio could take any value). Anyway in this case L'Hospital pertains
                              $$lim_{xto a}frac{f(x)}{g(x)}=lim_{xto a}frac{f'(x)}{g'(x)}.$$



                              Compare



                              enter image description here



                              and



                              enter image description here






                              share|cite|improve this answer











                              $endgroup$



                              If $f(a)=0$ while $g(a)ne0$ then



                              $$lim_{xto a}frac{f(x)}{g(x)}=0$$ because in the vicinity of $a$, the ratio get smaller and smaller.



                              If $f(a)=0$ while $g(a)=0$, you cannot say (because the ratio could take any value). Anyway in this case L'Hospital pertains
                              $$lim_{xto a}frac{f(x)}{g(x)}=lim_{xto a}frac{f'(x)}{g'(x)}.$$



                              Compare



                              enter image description here



                              and



                              enter image description here







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Jan 24 at 21:21

























                              answered Jan 24 at 21:14









                              Yves DaoustYves Daoust

                              130k676229




                              130k676229























                                  0












                                  $begingroup$


                                  What are Limits?



                                  Limits are the way to find the value of a function when we get an
                                  indeterminate form of a number (like $frac 0 0$ which cannot be
                                  represented in Real number Line Axis).



                                  When we plot such functions where there is an indeterminate form at a
                                  certain value of $x$, then we don't get holes or gaps (try yourself!).
                                  Why? because they are not defined at a point but everywhere near them,
                                  so, that point is indefinitely small.



                                  Limits are the mathematical way to say almost when it's too small, so,




                                  unless you get a number in the indeterminate form, limits are not useful in such cases. So, next time, just simply put the value and evaluate the function like just any other function ;)



                                  Here in your case when we plug the value of $x$, we simply get 0, that's what the limit of the function is!



                                  Note: It's also possible that Limit of a function is not defined at a point, but the direct value exists! think yourself for such examples :P :)






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$


                                    What are Limits?



                                    Limits are the way to find the value of a function when we get an
                                    indeterminate form of a number (like $frac 0 0$ which cannot be
                                    represented in Real number Line Axis).



                                    When we plot such functions where there is an indeterminate form at a
                                    certain value of $x$, then we don't get holes or gaps (try yourself!).
                                    Why? because they are not defined at a point but everywhere near them,
                                    so, that point is indefinitely small.



                                    Limits are the mathematical way to say almost when it's too small, so,




                                    unless you get a number in the indeterminate form, limits are not useful in such cases. So, next time, just simply put the value and evaluate the function like just any other function ;)



                                    Here in your case when we plug the value of $x$, we simply get 0, that's what the limit of the function is!



                                    Note: It's also possible that Limit of a function is not defined at a point, but the direct value exists! think yourself for such examples :P :)






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$


                                      What are Limits?



                                      Limits are the way to find the value of a function when we get an
                                      indeterminate form of a number (like $frac 0 0$ which cannot be
                                      represented in Real number Line Axis).



                                      When we plot such functions where there is an indeterminate form at a
                                      certain value of $x$, then we don't get holes or gaps (try yourself!).
                                      Why? because they are not defined at a point but everywhere near them,
                                      so, that point is indefinitely small.



                                      Limits are the mathematical way to say almost when it's too small, so,




                                      unless you get a number in the indeterminate form, limits are not useful in such cases. So, next time, just simply put the value and evaluate the function like just any other function ;)



                                      Here in your case when we plug the value of $x$, we simply get 0, that's what the limit of the function is!



                                      Note: It's also possible that Limit of a function is not defined at a point, but the direct value exists! think yourself for such examples :P :)






                                      share|cite|improve this answer









                                      $endgroup$




                                      What are Limits?



                                      Limits are the way to find the value of a function when we get an
                                      indeterminate form of a number (like $frac 0 0$ which cannot be
                                      represented in Real number Line Axis).



                                      When we plot such functions where there is an indeterminate form at a
                                      certain value of $x$, then we don't get holes or gaps (try yourself!).
                                      Why? because they are not defined at a point but everywhere near them,
                                      so, that point is indefinitely small.



                                      Limits are the mathematical way to say almost when it's too small, so,




                                      unless you get a number in the indeterminate form, limits are not useful in such cases. So, next time, just simply put the value and evaluate the function like just any other function ;)



                                      Here in your case when we plug the value of $x$, we simply get 0, that's what the limit of the function is!



                                      Note: It's also possible that Limit of a function is not defined at a point, but the direct value exists! think yourself for such examples :P :)







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 25 at 15:45









                                      Abhas Kumar SinhaAbhas Kumar Sinha

                                      304115




                                      304115






























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