Question

Find the limit:
x to infinity

xsin(2/x)

Answer 1

Break it up in parts lim x and lim sin 2/x. For sin 2/x, look in your notes for what happens when a number say a is over in finity. Therefore it becomes sin of that quantity.

Answer 2

however i am not getting the answer by applying the limit the following way:
|sin(2/x)|<= 1
so |xsin(2/x)|<=|x||sin2/x| triangle inequality
now if i apply limit,
lim |x|=infinity since it diverges, the smaller ones are bound to diverge. but in effect the answer is 2

Answer 3

How do you get 2, rsaad2?

Answer 4

can someone explain as to why? you use taylor polynomial, you get 2, but with comparison test you get infinity. confused ;S

Answer 5

Why are you going into higher mathematics? Why not Cal I limits?

Answer 6

as in?

Answer 7

like i said i am using simple limit, i am getting infinity.

Answer 8

Saw a similar example in my text book. Looking it up, I'll get back...

Answer 9

okay. here it is
lim x (2/x) {sin(2/x)/ (2/x)}
lim 2 lim {sin(2/x)/ (2/x)}
2*1
2

Answer 10

and the reason as to why sin(2/x)/(2/x) = 1 is because its similar to sin x/x when x-> 0
so lim sin x/x = 1.

Answer 11

Yeah, I was researching for my own learning. The proof is let u =sin 2x, then x = 2/u; limits changed to u goes to zero. It becomes limit as u goes to zero of (2 sin u)/u. Of course (sin u)/u is an identity equal to one. Therefore answer two. Good job rsaad2.