Left-hand limit at x = 0

Since we are approaching $0$ from the left, we use the piece of the function defined for $x \le 0$, which is $f(x) = x^2 - 1$.

Substituting $x = 0$ into this expression gives $0^2 - 1 = -1$. Therefore $\lim_{x \to 0^-} f(x) = -1$.

Since we are approaching $0$ from the right, we use the piece of the function defined for $x > 0$, which is $f(x) = 2x - 1$.

Substituting $x = 0$ into this expression gives $2(0) - 1 = -1$. Therefore $\lim_{x \to 0^+} f(x) = -1$.

We found $\lim_{x \to 0^-} f(x) = -1$ and $\lim_{x \to 0^+} f(x) = -1$.

Since both one-sided limits exist and are equal, the two-sided limit exists and equals their common value: $\lim_{x \to 0} f(x) = -1$.

No, it would not change. The right-hand limit as x approaches zero only depends on the values of the function for x close to zero, not far away. Since the near-zero behavior of the formula two x minus one is unchanged, the right-hand limit remains negative one. This reflects the core idea that limits depend only on nearby behavior, not on the function's behavior elsewhere.