Question

why does L'hospital's rule fail for evaluating this limit:
x tends to 0, [e^(1/x) - 1]/[e^(1/x) + 1]. The RHL is not equal to LHL clearly limit doesnt exist however LH rule yields the value of RHL(1). why??

Answer 1

I found this at matharticles.com on L'Hopitals rule that might shed some light on your problem.

http://www.maa.org/pubs/calc_articles/ma041.pdf

Good luck,
TTD

Answer 2

the function isn't continues at x=0 and even isn't defined at 0 so you can't calculate its derivative and use that in L'Hopital's rule at x=0 (its graph is could be shaped from tanhx, with 1/x=y substituion)

Answer 3

@arnp5
LHospitals do apply since....U will get inderterminant of the form \[\infty/\infty\]

Answer 4

and the answer is 1