Question

evaluate limx->infinity of xsin(1/x)

Answer 1

just put 1/x =y

Answer 2

i know its 1 from the graph but thats not a valid answer

Answer 3

what do you mean @hartnn

Answer 4

it is =1

Answer 5

but how do get 1

Answer 6

why do u say 1 is not valid?

Answer 7

ok, when x =1/y
y=1/x
when x->infinity , y -> 0
got this ?

Answer 8

no i need to get the answer. saying i used the graph is not a valid explanation

Answer 9

ohh, then yes, you the substitution that i suggested.

Answer 10

when x =1/y
y=1/x
when x->infinity , y -> 0
got this ?

Answer 11

no i dont get that

Answer 12

you know 1/infinity = 0 ?

Answer 13

yeah

Answer 14

so when x-> infinity
1/x -> 1/ infinity

but 1/x i put = y
so, y-> 1/infinity
so, y-> 0
now clear ?

Answer 15

so u said \[1/\infty=y\]?

Answer 16

no, i said, 1/infinity = 0
1/x =y

Answer 17

ok and then what

Answer 18

so we now have
lim y->0 sin y/y
right ?

Answer 19

wouldnt it be ysin(1/y)

Answer 20

it was x sin (1/x)
we changed x to 1/y
so, 1/x = y
so, x sin(1/x) changed to sin y/y
ok?

Answer 21

ok got it

Answer 22

so its sin(1)=1?

Answer 23

it is a standard limit formula that
lim x-> 0 sin x/ x = 1
so, also
lim y->0 sin y/y =1

Answer 24

but i am trying to find limx->infinity

Answer 25

didn't i explain how we changed that to y-> 0
?

Answer 26

no

Answer 27

"
so when x-> infinity
1/x -> 1/ infinity

but 1/x i put = y
so, y-> 1/infinity
so, y-> 0
now clear ?
"

Answer 28

He did. A few times

Answer 29

you should probably start from the top of the thread again and read the entire thing again

Answer 30

^

Answer 31

ok i got it thanks so much

Answer 32

i am glad to help :)
welcome ^_^