Evaluating $lim_{xto 2^+} (x-2)^{x^2-4}$












0














I am trying to evaluate this limit as follows,
$$lim_{xto 2^+} (x-2)^{x^2-4} = e^{lim_{xto 2^+} x^2-4 cdot ln(x-2)}$$



Aside, we can apply L'Hospital to the limit in the exponent by rearranging to form $frac{-infty}{infty}$:
$$lim_{xto 2^+} frac{ln(x-2)}{frac{1}{x^2-4}} = lim_{xto 2^+}frac{frac{1}{x-2}}{frac{2x}{(x^2-4)^2}}$$



At this point I'm kind of lost as to what I should do. I tried rearranging the fraction and then simplifying as follows and applying L'Hospital again since we have $frac{0}{0}$
$$=lim_{xto 2^+}frac{(x^2-4)^2}{(x-2)(2x)} = lim_{xto 2^+}frac{2(x^2-4)cdot2x}{4x-4} =frac{0}{4} =0 $$
I am supposed to get $1$ as the answer. I have a feeling I might be doing some sort of illegal manipulation, because I've somehow managed to make this limit equal to $4$ and $8$ as well using similiar techniques. What am I doing wrong?



Am I allowed to restructure the limit as I did after applying L'Hospital's rule once? If I keep it in the same fraction over another fraction form and apply L'Hospital's rule again, I get something even messier.






calculus limits derivatives







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  • math.stackexchange.com/questions/470952/…
    – lab bhattacharjee
    Dec 6 at 9:47
















0














I am trying to evaluate this limit as follows,
$$lim_{xto 2^+} (x-2)^{x^2-4} = e^{lim_{xto 2^+} x^2-4 cdot ln(x-2)}$$



Aside, we can apply L'Hospital to the limit in the exponent by rearranging to form $frac{-infty}{infty}$:
$$lim_{xto 2^+} frac{ln(x-2)}{frac{1}{x^2-4}} = lim_{xto 2^+}frac{frac{1}{x-2}}{frac{2x}{(x^2-4)^2}}$$



At this point I'm kind of lost as to what I should do. I tried rearranging the fraction and then simplifying as follows and applying L'Hospital again since we have $frac{0}{0}$
$$=lim_{xto 2^+}frac{(x^2-4)^2}{(x-2)(2x)} = lim_{xto 2^+}frac{2(x^2-4)cdot2x}{4x-4} =frac{0}{4} =0 $$
I am supposed to get $1$ as the answer. I have a feeling I might be doing some sort of illegal manipulation, because I've somehow managed to make this limit equal to $4$ and $8$ as well using similiar techniques. What am I doing wrong?



Am I allowed to restructure the limit as I did after applying L'Hospital's rule once? If I keep it in the same fraction over another fraction form and apply L'Hospital's rule again, I get something even messier.






calculus limits derivatives







share|cite|improve this question
























  • math.stackexchange.com/questions/470952/…
    – lab bhattacharjee
    Dec 6 at 9:47














0












0








0







I am trying to evaluate this limit as follows,
$$lim_{xto 2^+} (x-2)^{x^2-4} = e^{lim_{xto 2^+} x^2-4 cdot ln(x-2)}$$



Aside, we can apply L'Hospital to the limit in the exponent by rearranging to form $frac{-infty}{infty}$:
$$lim_{xto 2^+} frac{ln(x-2)}{frac{1}{x^2-4}} = lim_{xto 2^+}frac{frac{1}{x-2}}{frac{2x}{(x^2-4)^2}}$$



At this point I'm kind of lost as to what I should do. I tried rearranging the fraction and then simplifying as follows and applying L'Hospital again since we have $frac{0}{0}$
$$=lim_{xto 2^+}frac{(x^2-4)^2}{(x-2)(2x)} = lim_{xto 2^+}frac{2(x^2-4)cdot2x}{4x-4} =frac{0}{4} =0 $$
I am supposed to get $1$ as the answer. I have a feeling I might be doing some sort of illegal manipulation, because I've somehow managed to make this limit equal to $4$ and $8$ as well using similiar techniques. What am I doing wrong?



Am I allowed to restructure the limit as I did after applying L'Hospital's rule once? If I keep it in the same fraction over another fraction form and apply L'Hospital's rule again, I get something even messier.






calculus limits derivatives







share|cite|improve this question















I am trying to evaluate this limit as follows,
$$lim_{xto 2^+} (x-2)^{x^2-4} = e^{lim_{xto 2^+} x^2-4 cdot ln(x-2)}$$



Aside, we can apply L'Hospital to the limit in the exponent by rearranging to form $frac{-infty}{infty}$:
$$lim_{xto 2^+} frac{ln(x-2)}{frac{1}{x^2-4}} = lim_{xto 2^+}frac{frac{1}{x-2}}{frac{2x}{(x^2-4)^2}}$$



At this point I'm kind of lost as to what I should do. I tried rearranging the fraction and then simplifying as follows and applying L'Hospital again since we have $frac{0}{0}$
$$=lim_{xto 2^+}frac{(x^2-4)^2}{(x-2)(2x)} = lim_{xto 2^+}frac{2(x^2-4)cdot2x}{4x-4} =frac{0}{4} =0 $$
I am supposed to get $1$ as the answer. I have a feeling I might be doing some sort of illegal manipulation, because I've somehow managed to make this limit equal to $4$ and $8$ as well using similiar techniques. What am I doing wrong?



Am I allowed to restructure the limit as I did after applying L'Hospital's rule once? If I keep it in the same fraction over another fraction form and apply L'Hospital's rule again, I get something even messier.






calculus limits derivatives




calculus limits derivatives






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edited Dec 6 at 11:35









Tianlalu

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asked Dec 6 at 9:42









Danielle

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  • math.stackexchange.com/questions/470952/…
    – lab bhattacharjee
    Dec 6 at 9:47


















  • math.stackexchange.com/questions/470952/…
    – lab bhattacharjee
    Dec 6 at 9:47
















math.stackexchange.com/questions/470952/…
– lab bhattacharjee
Dec 6 at 9:47




math.stackexchange.com/questions/470952/…
– lab bhattacharjee
Dec 6 at 9:47










4 Answers
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You did alright, you just forget the exponential outside. $exp(0)=1$.



Alternatively, note that



$$lim_{x to 2^+}frac{(x^2-4)^2}{(x-2)(2x)}=lim_{x to 2^+}frac{(x-2)^2(x+2)^2}{(x-2)(2x)}=lim_{x to 2^+}frac{(x-2)(x+2)^2}{(2x)}=0$$



then remember to take exponential.






share|cite|improve this answer





























    0














    You may substitute $t = x-2$ and consider $t rightarrow 0^+$ and use





    • $t^t stackrel{t to 0^+}{longrightarrow} 1$ and note that

    • $x^2-4 = (x-2)(x+2)$


    $$(x-2)^{x^2-4} = left( t^tright)^{t+4} stackrel{t to 0^+}{longrightarrow} 1^4 = 1$$






    share|cite|improve this answer





























      0














      Let $ y=(x-2)^{x^2-4}$, then $ log y=(x^2-4) log (x-2)=frac{log(x-2)}{frac{1}{x^2-4}}$.



      Now $ lim_{x to 2^{+}} log y= lim_{x to 2^{+}} frac{log(x-2)}{frac{1}{x^2-4}}= left( frac{infty}{infty} right) form $



      By L'Hospital's rule



      $ lim_{x to 2^{+}} log y= lim_{x to 2^{+}} frac{(x^2-4)^2}{-2x(x-2)}=lim_{x to 2^{+}} frac{(x-2)(x+2)^2}{-2x}=0 $



      $ Rightarrow lim_{x to 2^{+}} log y=0=log 1=lim_{x to 2^{+}} log 1 \ Rightarrow lim_{x to 2^{+}} y=1 \ Rightarrow lim_{x to 2^{+}}(x-2)^{x^2-4}=1 $






      share|cite|improve this answer





























        0














        We have that



        $$(x-2)^{x^2-4}=e^{(x^2-4)log(x-2)} to 1$$



        indeed by standard limits by $t=x-2 to 0^+$



        $$(x^2-4)log(x-2)=(t+4)cdot tlog t to 4cdot0=0$$



        indeed by $y=frac1t to infty$



        $$tlog t=frac1y log left(frac1yright)=-frac{log y}{y} to 0$$



        and the last we can refer to the related




        • Compute $lim_{x to infty} frac{log(x)}{x^a}$






        share|cite|improve this answer























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          4 Answers
          4






          active

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          4 Answers
          4






          active

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          active

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          active

          oldest

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          2














          You did alright, you just forget the exponential outside. $exp(0)=1$.



          Alternatively, note that



          $$lim_{x to 2^+}frac{(x^2-4)^2}{(x-2)(2x)}=lim_{x to 2^+}frac{(x-2)^2(x+2)^2}{(x-2)(2x)}=lim_{x to 2^+}frac{(x-2)(x+2)^2}{(2x)}=0$$



          then remember to take exponential.






          share|cite|improve this answer


























            2














            You did alright, you just forget the exponential outside. $exp(0)=1$.



            Alternatively, note that



            $$lim_{x to 2^+}frac{(x^2-4)^2}{(x-2)(2x)}=lim_{x to 2^+}frac{(x-2)^2(x+2)^2}{(x-2)(2x)}=lim_{x to 2^+}frac{(x-2)(x+2)^2}{(2x)}=0$$



            then remember to take exponential.






            share|cite|improve this answer
























              2












              2








              2






              You did alright, you just forget the exponential outside. $exp(0)=1$.



              Alternatively, note that



              $$lim_{x to 2^+}frac{(x^2-4)^2}{(x-2)(2x)}=lim_{x to 2^+}frac{(x-2)^2(x+2)^2}{(x-2)(2x)}=lim_{x to 2^+}frac{(x-2)(x+2)^2}{(2x)}=0$$



              then remember to take exponential.






              share|cite|improve this answer












              You did alright, you just forget the exponential outside. $exp(0)=1$.



              Alternatively, note that



              $$lim_{x to 2^+}frac{(x^2-4)^2}{(x-2)(2x)}=lim_{x to 2^+}frac{(x-2)^2(x+2)^2}{(x-2)(2x)}=lim_{x to 2^+}frac{(x-2)(x+2)^2}{(2x)}=0$$



              then remember to take exponential.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 6 at 9:47









              Siong Thye Goh

              98.3k1463116




              98.3k1463116























                  0














                  You may substitute $t = x-2$ and consider $t rightarrow 0^+$ and use





                  • $t^t stackrel{t to 0^+}{longrightarrow} 1$ and note that

                  • $x^2-4 = (x-2)(x+2)$


                  $$(x-2)^{x^2-4} = left( t^tright)^{t+4} stackrel{t to 0^+}{longrightarrow} 1^4 = 1$$






                  share|cite|improve this answer


























                    0














                    You may substitute $t = x-2$ and consider $t rightarrow 0^+$ and use





                    • $t^t stackrel{t to 0^+}{longrightarrow} 1$ and note that

                    • $x^2-4 = (x-2)(x+2)$


                    $$(x-2)^{x^2-4} = left( t^tright)^{t+4} stackrel{t to 0^+}{longrightarrow} 1^4 = 1$$






                    share|cite|improve this answer
























                      0












                      0








                      0






                      You may substitute $t = x-2$ and consider $t rightarrow 0^+$ and use





                      • $t^t stackrel{t to 0^+}{longrightarrow} 1$ and note that

                      • $x^2-4 = (x-2)(x+2)$


                      $$(x-2)^{x^2-4} = left( t^tright)^{t+4} stackrel{t to 0^+}{longrightarrow} 1^4 = 1$$






                      share|cite|improve this answer












                      You may substitute $t = x-2$ and consider $t rightarrow 0^+$ and use





                      • $t^t stackrel{t to 0^+}{longrightarrow} 1$ and note that

                      • $x^2-4 = (x-2)(x+2)$


                      $$(x-2)^{x^2-4} = left( t^tright)^{t+4} stackrel{t to 0^+}{longrightarrow} 1^4 = 1$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 6 at 9:50









                      trancelocation

                      8,9651521




                      8,9651521























                          0














                          Let $ y=(x-2)^{x^2-4}$, then $ log y=(x^2-4) log (x-2)=frac{log(x-2)}{frac{1}{x^2-4}}$.



                          Now $ lim_{x to 2^{+}} log y= lim_{x to 2^{+}} frac{log(x-2)}{frac{1}{x^2-4}}= left( frac{infty}{infty} right) form $



                          By L'Hospital's rule



                          $ lim_{x to 2^{+}} log y= lim_{x to 2^{+}} frac{(x^2-4)^2}{-2x(x-2)}=lim_{x to 2^{+}} frac{(x-2)(x+2)^2}{-2x}=0 $



                          $ Rightarrow lim_{x to 2^{+}} log y=0=log 1=lim_{x to 2^{+}} log 1 \ Rightarrow lim_{x to 2^{+}} y=1 \ Rightarrow lim_{x to 2^{+}}(x-2)^{x^2-4}=1 $






                          share|cite|improve this answer


























                            0














                            Let $ y=(x-2)^{x^2-4}$, then $ log y=(x^2-4) log (x-2)=frac{log(x-2)}{frac{1}{x^2-4}}$.



                            Now $ lim_{x to 2^{+}} log y= lim_{x to 2^{+}} frac{log(x-2)}{frac{1}{x^2-4}}= left( frac{infty}{infty} right) form $



                            By L'Hospital's rule



                            $ lim_{x to 2^{+}} log y= lim_{x to 2^{+}} frac{(x^2-4)^2}{-2x(x-2)}=lim_{x to 2^{+}} frac{(x-2)(x+2)^2}{-2x}=0 $



                            $ Rightarrow lim_{x to 2^{+}} log y=0=log 1=lim_{x to 2^{+}} log 1 \ Rightarrow lim_{x to 2^{+}} y=1 \ Rightarrow lim_{x to 2^{+}}(x-2)^{x^2-4}=1 $






                            share|cite|improve this answer
























                              0












                              0








                              0






                              Let $ y=(x-2)^{x^2-4}$, then $ log y=(x^2-4) log (x-2)=frac{log(x-2)}{frac{1}{x^2-4}}$.



                              Now $ lim_{x to 2^{+}} log y= lim_{x to 2^{+}} frac{log(x-2)}{frac{1}{x^2-4}}= left( frac{infty}{infty} right) form $



                              By L'Hospital's rule



                              $ lim_{x to 2^{+}} log y= lim_{x to 2^{+}} frac{(x^2-4)^2}{-2x(x-2)}=lim_{x to 2^{+}} frac{(x-2)(x+2)^2}{-2x}=0 $



                              $ Rightarrow lim_{x to 2^{+}} log y=0=log 1=lim_{x to 2^{+}} log 1 \ Rightarrow lim_{x to 2^{+}} y=1 \ Rightarrow lim_{x to 2^{+}}(x-2)^{x^2-4}=1 $






                              share|cite|improve this answer












                              Let $ y=(x-2)^{x^2-4}$, then $ log y=(x^2-4) log (x-2)=frac{log(x-2)}{frac{1}{x^2-4}}$.



                              Now $ lim_{x to 2^{+}} log y= lim_{x to 2^{+}} frac{log(x-2)}{frac{1}{x^2-4}}= left( frac{infty}{infty} right) form $



                              By L'Hospital's rule



                              $ lim_{x to 2^{+}} log y= lim_{x to 2^{+}} frac{(x^2-4)^2}{-2x(x-2)}=lim_{x to 2^{+}} frac{(x-2)(x+2)^2}{-2x}=0 $



                              $ Rightarrow lim_{x to 2^{+}} log y=0=log 1=lim_{x to 2^{+}} log 1 \ Rightarrow lim_{x to 2^{+}} y=1 \ Rightarrow lim_{x to 2^{+}}(x-2)^{x^2-4}=1 $







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 6 at 10:28









                              M. A. SARKAR

                              2,1271619




                              2,1271619























                                  0














                                  We have that



                                  $$(x-2)^{x^2-4}=e^{(x^2-4)log(x-2)} to 1$$



                                  indeed by standard limits by $t=x-2 to 0^+$



                                  $$(x^2-4)log(x-2)=(t+4)cdot tlog t to 4cdot0=0$$



                                  indeed by $y=frac1t to infty$



                                  $$tlog t=frac1y log left(frac1yright)=-frac{log y}{y} to 0$$



                                  and the last we can refer to the related




                                  • Compute $lim_{x to infty} frac{log(x)}{x^a}$






                                  share|cite|improve this answer




























                                    0














                                    We have that



                                    $$(x-2)^{x^2-4}=e^{(x^2-4)log(x-2)} to 1$$



                                    indeed by standard limits by $t=x-2 to 0^+$



                                    $$(x^2-4)log(x-2)=(t+4)cdot tlog t to 4cdot0=0$$



                                    indeed by $y=frac1t to infty$



                                    $$tlog t=frac1y log left(frac1yright)=-frac{log y}{y} to 0$$



                                    and the last we can refer to the related




                                    • Compute $lim_{x to infty} frac{log(x)}{x^a}$






                                    share|cite|improve this answer


























                                      0












                                      0








                                      0






                                      We have that



                                      $$(x-2)^{x^2-4}=e^{(x^2-4)log(x-2)} to 1$$



                                      indeed by standard limits by $t=x-2 to 0^+$



                                      $$(x^2-4)log(x-2)=(t+4)cdot tlog t to 4cdot0=0$$



                                      indeed by $y=frac1t to infty$



                                      $$tlog t=frac1y log left(frac1yright)=-frac{log y}{y} to 0$$



                                      and the last we can refer to the related




                                      • Compute $lim_{x to infty} frac{log(x)}{x^a}$






                                      share|cite|improve this answer














                                      We have that



                                      $$(x-2)^{x^2-4}=e^{(x^2-4)log(x-2)} to 1$$



                                      indeed by standard limits by $t=x-2 to 0^+$



                                      $$(x^2-4)log(x-2)=(t+4)cdot tlog t to 4cdot0=0$$



                                      indeed by $y=frac1t to infty$



                                      $$tlog t=frac1y log left(frac1yright)=-frac{log y}{y} to 0$$



                                      and the last we can refer to the related




                                      • Compute $lim_{x to infty} frac{log(x)}{x^a}$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Dec 6 at 11:59

























                                      answered Dec 6 at 10:07









                                      gimusi

                                      1




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