Question

Question #c69a9

Answer 1

The limit:

`lim_(x->0) (1-3x)^3/x`

does not exist.

Explanation:

The numerator is continuous for `x = 0`, so we have:

`lim_(x->0^+) (1-3x)^3/x = lim_(x->0^+) (1-3x)^3 * lim_(x->0^+) 1/x = 1*+oo =+oo`

`lim_(x->0^-) (1-3x)^3/x = lim_(x->0^-) (1-3x)^3 * lim_(x->0^-) 1/x = 1*-oo =-oo`

Then we can see that the left and right limits are different which means that the function does not have a limit for `x->0`