Evaluate the ff limits, if they exist (i) lim x->1 (x8 - 1)/(x10 - 1) (ii) lim x->3 (x2 - 9)/(x2 + 2x - 3)

If we substitute $x=1$ into the expression, we get $\frac{1^8 - 1}{1^{10} - 1} = \frac{0}{0}$, which is an indeterminate form. This means we can apply L'Hôpital's Rule.

Take the derivative of the numerator and the derivative of the denominator with respect to $x$.
Derivative of the numerator: $\frac{d}{dx}(x^8 - 1) = 8x^7$
Derivative of the denominator: $\frac{d}{dx}(x^{10} - 1) = 10x^9$

Substitute $x=1$:

The limit exists and is:

If we substitute $x=-3$ into the expression, we get $\frac{(-3)^2 - 9}{(-3)^2 + 2(-3) - 3} = \frac{9 - 9}{9 - 6 - 3} = \frac{0}{0}$, which is an indeterminate form. We can apply L'Hôpital's Rule or factor the expression. Let's use factoring for this one.

Numerator: $x^2 - 9 = (x-3)(x+3)$
Denominator: $x^2 + 2x - 3 = (x+3)(x-1)$

Since $x \to -3$, $x \neq -3$, so we can cancel the $(x+3)$ term.

Substitute $x=-3$:

The limit exists and is: