The function IS continuous at x=0 but you need to show that $\lim_{x\to 0} f(x)= 1$ to show THAT so continuity cannot be used to do this exercise.
Albert, you state that $-1\le x^2- 2\le 1$. That is the same as $1\le x^2\le 3$. If x is close to 0 that is NOT true! You also have $\lim_{x\to 0} -x^2= 1$ and $\lim_{x\to 0} x^2= 1$. Those are certainly not true! Did you mean "= 0"?