Question

How can we find the lim (when (x,y)->(0,0) ) of x* sin(1/x^2) ?

Answer 1

for x->0, Sin(1/x^2) will always have value 0<= sin(1/x^2) <=1 , since x->0, x is not actually zero.

so, x*some no = 0 so we have limit 0

Answer 2

Thank you!

Answer 3

you are welcome.