How can I find this limit? L'hopital's rule is never-ending












0












$begingroup$


I have the following limit
$$
lim_{xtoinfty}frac{-7^x}{(ln11-ln7)cdot11^x}$$
Graphing gives me $0$ but plugging it in gives infinity over infinity.



I then tried L'Hôpital's rule and it seems to fall into a never ending loop. I can't seem to simplify this - what would be the best way to find the limit?



Edit: would like to clarify I want to know how to find it algebraically






calculus limits







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  • $begingroup$
    Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways.
    $endgroup$
    – egreg
    Jan 12 at 23:53












  • $begingroup$
    @egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment)
    $endgroup$
    – Andrew Wang
    Jan 13 at 2:25
















0












$begingroup$


I have the following limit
$$
lim_{xtoinfty}frac{-7^x}{(ln11-ln7)cdot11^x}$$
Graphing gives me $0$ but plugging it in gives infinity over infinity.



I then tried L'Hôpital's rule and it seems to fall into a never ending loop. I can't seem to simplify this - what would be the best way to find the limit?



Edit: would like to clarify I want to know how to find it algebraically






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways.
    $endgroup$
    – egreg
    Jan 12 at 23:53












  • $begingroup$
    @egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment)
    $endgroup$
    – Andrew Wang
    Jan 13 at 2:25














0












0








0





$begingroup$


I have the following limit
$$
lim_{xtoinfty}frac{-7^x}{(ln11-ln7)cdot11^x}$$
Graphing gives me $0$ but plugging it in gives infinity over infinity.



I then tried L'Hôpital's rule and it seems to fall into a never ending loop. I can't seem to simplify this - what would be the best way to find the limit?



Edit: would like to clarify I want to know how to find it algebraically






calculus limits







share|cite|improve this question











$endgroup$




I have the following limit
$$
lim_{xtoinfty}frac{-7^x}{(ln11-ln7)cdot11^x}$$
Graphing gives me $0$ but plugging it in gives infinity over infinity.



I then tried L'Hôpital's rule and it seems to fall into a never ending loop. I can't seem to simplify this - what would be the best way to find the limit?



Edit: would like to clarify I want to know how to find it algebraically






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 23:46









egreg

186k1486208




186k1486208










asked Jan 12 at 23:18









Andrew WangAndrew Wang

415




415












  • $begingroup$
    Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways.
    $endgroup$
    – egreg
    Jan 12 at 23:53












  • $begingroup$
    @egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment)
    $endgroup$
    – Andrew Wang
    Jan 13 at 2:25


















  • $begingroup$
    Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways.
    $endgroup$
    – egreg
    Jan 12 at 23:53












  • $begingroup$
    @egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment)
    $endgroup$
    – Andrew Wang
    Jan 13 at 2:25
















$begingroup$
Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways.
$endgroup$
– egreg
Jan 12 at 23:53






$begingroup$
Is this the real exercise or did you get at this by applying l'Hôpital to something else? In the latter case it's quite likely that there are better ways.
$endgroup$
– egreg
Jan 12 at 23:53














$begingroup$
@egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment)
$endgroup$
– Andrew Wang
Jan 13 at 2:25




$begingroup$
@egreg No, it was part of an exercise in imperfect integrals (I don't recall the details of the questions at the moment)
$endgroup$
– Andrew Wang
Jan 13 at 2:25










2 Answers
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$begingroup$

You can write the expression as $-left(ln frac{11}{7}right)^{-1} left(frac{1}{(11/7)}right)^x$, which goes to zero as $x rightarrow infty$






share|cite|improve this answer











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    0












    $begingroup$

    Rewrite $$frac{-7^x}{(ln11-ln7),11^x}=- frac{exp(x log(7))}{log left(frac{11}{7}right) exp(x log(11))}=-frac{exp(x log(7)),exp(-x log(11))}{log left(frac{11}{7}right)}=-frac{exp(-x (log(11)-log(7)))}{log left(frac{11}{7}right)}=-frac{exp(-x log left(frac{11}{7}right))}{log left(frac{11}{7}right)}$$ and take ito account the fact that $log left(frac{11}{7}right)>0$






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      2 Answers
      2






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      2 Answers
      2






      active

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      active

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      active

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      4












      $begingroup$

      You can write the expression as $-left(ln frac{11}{7}right)^{-1} left(frac{1}{(11/7)}right)^x$, which goes to zero as $x rightarrow infty$






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        You can write the expression as $-left(ln frac{11}{7}right)^{-1} left(frac{1}{(11/7)}right)^x$, which goes to zero as $x rightarrow infty$






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          You can write the expression as $-left(ln frac{11}{7}right)^{-1} left(frac{1}{(11/7)}right)^x$, which goes to zero as $x rightarrow infty$






          share|cite|improve this answer











          $endgroup$



          You can write the expression as $-left(ln frac{11}{7}right)^{-1} left(frac{1}{(11/7)}right)^x$, which goes to zero as $x rightarrow infty$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 12 at 23:40

























          answered Jan 12 at 23:23









          pendermathpendermath

          57612




          57612























              0












              $begingroup$

              Rewrite $$frac{-7^x}{(ln11-ln7),11^x}=- frac{exp(x log(7))}{log left(frac{11}{7}right) exp(x log(11))}=-frac{exp(x log(7)),exp(-x log(11))}{log left(frac{11}{7}right)}=-frac{exp(-x (log(11)-log(7)))}{log left(frac{11}{7}right)}=-frac{exp(-x log left(frac{11}{7}right))}{log left(frac{11}{7}right)}$$ and take ito account the fact that $log left(frac{11}{7}right)>0$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Rewrite $$frac{-7^x}{(ln11-ln7),11^x}=- frac{exp(x log(7))}{log left(frac{11}{7}right) exp(x log(11))}=-frac{exp(x log(7)),exp(-x log(11))}{log left(frac{11}{7}right)}=-frac{exp(-x (log(11)-log(7)))}{log left(frac{11}{7}right)}=-frac{exp(-x log left(frac{11}{7}right))}{log left(frac{11}{7}right)}$$ and take ito account the fact that $log left(frac{11}{7}right)>0$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Rewrite $$frac{-7^x}{(ln11-ln7),11^x}=- frac{exp(x log(7))}{log left(frac{11}{7}right) exp(x log(11))}=-frac{exp(x log(7)),exp(-x log(11))}{log left(frac{11}{7}right)}=-frac{exp(-x (log(11)-log(7)))}{log left(frac{11}{7}right)}=-frac{exp(-x log left(frac{11}{7}right))}{log left(frac{11}{7}right)}$$ and take ito account the fact that $log left(frac{11}{7}right)>0$






                  share|cite|improve this answer









                  $endgroup$



                  Rewrite $$frac{-7^x}{(ln11-ln7),11^x}=- frac{exp(x log(7))}{log left(frac{11}{7}right) exp(x log(11))}=-frac{exp(x log(7)),exp(-x log(11))}{log left(frac{11}{7}right)}=-frac{exp(-x (log(11)-log(7)))}{log left(frac{11}{7}right)}=-frac{exp(-x log left(frac{11}{7}right))}{log left(frac{11}{7}right)}$$ and take ito account the fact that $log left(frac{11}{7}right)>0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 13 at 3:58









                  Claude LeiboviciClaude Leibovici

                  126k1158135




                  126k1158135






























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