Question

lim xsin(e^-x) [x tends to infinity]

Answer 1

Did you try to write \[\large x \sin(e^{-x}) = \frac{\sin(e^{-x})}{\frac{1}{x}} \] Where \(\sin\) is a continuous mapping, hence you will obtain the form \(0/0\) which can often elegantly be solved using L'Hospitals' rule. Here you might run into trouble though, hence I might recommend to you to use a taylor approach.

Answer 2

The answer is given as 0.But,i could not get the answer.

Answer 3

You should let us know then what methods you are familiar with.

Answer 4

I am trying to fall the problem in lim x->0 sinx/x=1

Answer 5

x sin(e^-x) = x sin(e^-x) / [ e^x e^(-x)] {note: e^x e^-x =1)

x/e^x * sin(e^-x)/e^-x

as x approaches infinity, e^-x approaches 0, and we know
lim{u->0} of sin(u)/u = 1

x/e^x approaches 0 when x approaches infinity,

so you have
x/e^x * sin(e^-x)/e^-x = 0 * 1 = 0

Answer 6

Why x/e^x approaches 0 when x approaches infinity?

Answer 7

Yes,i got it.Thanks@sourwing.