what is the limit of (x^3)/|x| as x approaches zero from the right?

Question

And as x approaches zero from the left.
Thanks in advance.

Answer 1

When you plug in 0, you get 0/0, so use l'Hopital's rule. Taking the derivative of the top and bottom gives (3x^2)/1. This limit approaches 0, so the original limit is 0.

You could also say that since x is approaching 0 from the right, then x is going to be negative, so |x| = -x. This gives you (x^3)/(-x), or -x^2 as the expression. The limit as this approaches 0 is obviously 0.

Answer 2

Since x is approaching 0 from the right the function is just x^3/x = x^2, so you can plug in x=0 to get the limit is 0.

Answer 3

As x approaches 0 from the left x is negative so IxI=-x and your

expression is x^3/(-x) = -x^2 which has limit 0 approaching it by negative values
If x approaches 0 from the right IxI = x
you have x^3/x=x^2 which approaches 0 with positive values

Answer 4

I think it must be zero in both cases. When x>0, |x|=x, so it's x^2 when x is not zero, but it approaches zero.

from the left, |x| = -x, so the expression is -x^2, which still approaches 0