Question

lim x->0 ( xsin(1/x) )
well i must be finding the value but im not sure wether it even exists!!

Answer 1

this limit is defined in its current form, you can simply substitute

Answer 2

i get 0*sin(infinity) is that even acceptable
can i just put zero as the answer??

Answer 3

why would you put infinity inside the sine term?

Answer 4

OOOOOO sorry i put the question wrong :P
its sin(1/x)

Answer 5

ah, in this case try making the substitution x=1/t

Answer 6

i edited it now :P

Answer 7

x=1/t

Answer 8

ohh thats a nice idea im tryin it now..

Answer 9

let me know how it works out for you :)

Answer 10

that means t is tending to infinity right??

Answer 11

right, actually you can still see the answer without the substitution as easily using the squeeze theorem

Answer 12

remember that no matter the argument, sin(x) has an upper and lower bound

Answer 13

what theoram o.O
sorry never heard of it :P

Answer 14

ya it is always between 1 and -1

Answer 15

right, so -x<= xsin(x) <=x

Answer 16

okay..... then what??

Answer 17

well, what is the limit x->0 of -x ?

Answer 18

zero??

Answer 19

right, and lim x->0 of x ?

Answer 20

zero again

Answer 21

OHH so the middle thingy is zero so the answer is zero too ??

Answer 22

exactly and since xsin(x) is always between those other two functions, it's limit can be seen by "squeezing" the other two functions together, since they have the same limit

Answer 23

OOOOOOOO so thats whats callified as squeez theorem eh??

Answer 24

It is worth remembering:

0*(bounded function) gives you zero in case of limits.

Answer 25

Thanks a lot ..... TuringTest... (*im sure thats not your real name though*) :P

Answer 26

thanks AravindG

Answer 27

hence the name, yeah
more formally stated as if g(x)<=f(x)<=h(x), and lim x->a g(x) = lim x->a h(x) = L
then lim x->a f(x)=L

Answer 28

Thanks a lot :P

Answer 29

welcome!

Answer 30

As other posters suggested
\[
\left | x \sin\left( \frac 1 x\right ) \right|\le |x|

\]