Why is the limit as x--> 0 for x^2 sin (1/x) = 0?

Question

Is there any way to do this using limit laws and not a table?

Answer 1

Since -1 <= sin(1/x) <= 1 for all x not 0, we have -x^2 <= sin(1/x) <= x^2 for x NE 0. Therefore, lim -x^2 <= lim sin(1/x) <= lim x^2. Since both -x^2 and x^2 have limit 0, the function squeezed in the middle must also have limit 0.

Answer 2

well if you put x=0 in your eqaution you well get
0^2 sin (1/0)=0
because whn we multiply 0 with any number we get result 0
one more thing
did you now if devide
1 / 0=infinty

Answer 3

I think that should be infinite since x approaches 0, so 1/0 = which is infinite

Answer 4

Pluge into the expression x = +10^-6 and x = -10^-6 and you will see that it converges on zero from above and below.