if we consider 1/exp(sinx) and take the limit as x--> infinity i think it's intuitively clear that the limit doesn't exist because -1<=sinx<=1. But if we multiply top and bottom by x (which doesn't change 1/exp(sinx)) and apply L'Hopital's rule we can argue that the limit is 0. Try differentiating top and bottom and look at the result. there is a factor of x in the denominator which will take the denominator to infinity while the numerator is 1.
can someone resolve this inconsistency?
i think what's important here is to be careful when quickly assuming the expression goes to 0 because the denominator goes to infinity while the numerator is a fixed constant. i made the mistake of not noticing that the denominator oscillates between infinity and - infinity. so in effect the limit doesn't exist and thus invoking L'Hopital's rule a mistake.
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!!