Thread: L'Hopital's rule question

yes. i had created an inconsistency. i knew 1/e^sinx had no limit. i multiplied by 1 (x/x) and then applied L'Hopital's rule, INCORRECTLY, and then had an inconsistency because applying the rule was giving me that the limit was 0. it was this inconsistency i was trying to resolve.

Originally Posted by kkoutsothodoros

originally when i posted this thread i knew 1/e^sinx had no limit as x-->infinity. what i did was multiply top and bottom by x. i then used L'Hopital's rule and incorrectly showed it's limit was 0. i couldn't resolve the inconsistency. i assumed after taking derivative of top and bottom of x/xe^sinx that the derivative of the bottom (e^sinx+xcosxe^sinx) goes to infinity. it doesn't because it oscillates between positive and negative infininty. i was only seeing the positive!!

The L'Hopital's rule can only be directly used for two types: type "0/0" and type "infinity/infinity"

Originally Posted by curvature

The L'Hopital's rule can only be directly used for two types: type "0/0" and type "infinity/infinity"

the latter condition was fulfilled when he multiplied by x/x

yes, so that's why i applied it to x/xe^sinx. i believe the limit as x-->infininit for both x and xe^sinx is infininty.

Originally Posted by kkoutsothodoros

yes. i had created an inconsistency. i knew 1/e^sinx had no limit. i multiplied by 1 (x/x) and then applied L'Hopital's rule, INCORRECTLY, and then had an inconsistency because applying the rule was giving me that the limit was 0. it was this inconsistency i was trying to resolve.

L'Hopital rule cannot be used for x/(x*exp(sinx)) since the denominator oscillates between e*x and x/e (see the graph below).

Originally Posted by Jhevon

the latter condition was fulfilled when he multiplied by x/x

Incorrect. see the last graph.

yes but if you consider the lower bound line in the graph it is going to infinity and therefore xe^sinx is going to infininty.

Originally Posted by kkoutsothodoros

yes but if you consider the lower bound line in the graph it is going to infinity and therefore xe^sinx is going to infininty.

Oh yes thanks

all in all i think this was a good problem to get the wheels turning in my head!

Originally Posted by kkoutsothodoros

yes but if you consider the lower bound line in the graph it is going to infinity and therefore xe^sinx is going to infininty.

However, after take derivatives, the limit of the new fraction dose not exist. So L'Hopital's rule cannot be used. The conditions of using the rule are not satisfied.

Originally Posted by curvature

However, after take derivatives, the limit of the new fraction dose not exist. So L'Hopital's rule cannot be used. The conditions of using the rule are not satisfied.

The rule can be used, it just reaffirms our original thought. the limit still doesn't exist (which is what we originally thought but wanted verification)

for some reason the poster wanted to convince himself that the infinite limit of does not exist

yes, if you draw one of those graphs again (could you post one?) for te denominator (e^sinx(1+xcosx)) after applying L'Hoptal's rule it will oscillate between infinity and -infinity so the limit doesn't exist (it is not infinity or -infininty) and so L'Hopital's rule, although used correctly, is not giving me an answer and so the incinsistency i was concerned with is resolved.

or what Jhevon said!

Originally Posted by kkoutsothodoros

yes, if you draw one of those graphs again (could you post one?) for te denominator (e^sinx(1+xcosx)) after applying L'Hoptal's rule it will oscillate between infinity and -infinity

Yes you are right.