Question

find lim xsin(1/x).
x->0

Answer 1

I think it must be as x approaches to infinity (not zero). If so the answer is one.

Answer 2

If x is approaching 0, and you're multiplying sine by x, then you're multiplying by zero, so the limit goes to 0.

Answer 3

you could rewrite it, using n = 1/x
then
\[\lim_{n \rightarrow \infty}\frac{\sin n}{n} = 0\]

Answer 4

Or more convincingly, as x->0, 1/x gets very large, but the sine of a very large number is still something between -1 and 1, and you are multiplying that by a very small number, so the limit approaches a very small number, i.e. zero.

Answer 5

can u please show how you got 0 b/c i got 1 as my answer... ???

Answer 6

we have...

Answer 7

\[\lim_{x \rightarrow 0} 5x = 0\]
right?
well think of sin(1/x) as just some constant as @CliffSedge posted above

Answer 8

oh well this how i got the answer
i know that lim sinx/x (x->0) equals to 1
so for lim xsin(1/x),
x->0
i did this lim (sin(1/x)/ (1/x)) =1
x->0

Answer 9

Yeah, the range of the sine function is a number between -1 and 1, so it's just some small number that is being multiplied by x and x is going to zero, so . .

Answer 10

If you change x to 1/x like that, then you have to change the limit.

Answer 11

Look at what Dumbcow first posted.

Answer 12

x and 1/x go to different limits as x goes to zero.

Answer 13

This really just breaks down to 'anything times zero is zero.'

Answer 14

but dumbcow used x->oo instead of x->0
sorry im still kinda confused...

Answer 15

sin(1/x) is just some 'anything.'

Answer 16

dumbcow changed x->0 to x->∞ because he changed sin(x)/x to sin(1/x)/(1/x).

Answer 17

as x->0, 1/x->∞

Answer 18

(Sort of)

Answer 19

it's actually ±∞, but the point is the same.

Answer 20

I think following my line of reasoning and working it out logically makes more sense and is easier than trying to fit it into some formula.

Answer 21

I'll repeat my argument for clarity.
Given:
\[\lim_{x \rightarrow 0} \space x \cdot \sin(1/x)\]
Taken one at a time,
\[\lim_{x \rightarrow 0} \space x = 0.\]
\[\lim_{x \rightarrow 0} \space \sin(1/x) = y:y=some \space number.\]
That 'some number,' y is between -1 and 1 because that is the range of sin(Θ).
So you're multiplying some number, y by zero.