Question

lim x^2 sin^2( 1\x )
x ->0

Answer 1

take a look at these images...

Answer 2

Try writing t = 1/x and substituting in. I think it should work fine.

Answer 3

that makes sense

Answer 4

That's t = 1/x, x^2 = 1/t^2. The limit will go to 0.

Answer 5

that's zero.

Answer 6

Because you can change the indeterminate form, i.e., you can write it as:\[\lim_{x \rightarrow 0} x^2 \sin^2(\frac{1}{x}) = \lim_{t \rightarrow \infty} \frac{\sin^2(t)}{t^2}\]Alternatively, you can prove it by the squeeze theorem as shown on these slides.

Answer 7

when sin ( 1 \ x ) doesn't have limit as x -> 0 then how this expression has?

Answer 8

Generally, limits are quite counter-intuitive, I agree.

Answer 9

it has some finite value

sin (any number) is always between -1 and +1

Answer 10

then we can change lim x->0 sin( 1/ x ) to lim x -> 0 sin( t ) where t = 1/x

Answer 11

0 * finite number between +1 and -1 is zero.

Answer 12

No, notice that I changed the limit also. It has to be of the form 0*infinity. But I think the squeeze theorem is more intuitive for this. sin has a boundary, while x^2 will go to 0.

Answer 13

@bmb ..ohhh.. i didn't pay attention to that....

Answer 14

first of all, let's see the graph of sin(1/x)