Question

closed.

Answer 1

i've been studying limits lately and dont find them easy
but if you apply l'hopitals rule you get

-e^-x / 1

= -1 / e^x

and limit of this as x --> infinity is 0

Answer 2

yea - thats what wolframalpha gives

Answer 3

yes i think so so thats 0 right?

Answer 4

so l'hopitals looks like its unnecessary

Answer 5

as i say limits are not my strong point ...

Answer 6

no it 0 / infinity ??? - i'm not sure to be honest
the limit is definitely 0 though

Answer 7

yeah, you cannot apply L'Hopitals, and the limit is 0

Answer 8

substitute x=1/y

Answer 9

lim y-> 0 y.e^(-1/y)

now directly put y=0
e^(-1/0) = e^(-infinity) = 0