Evaluate the following limits: (a) lim x->2 (x3 - 8)/(3x2 - 2x), (b) lim x->0 (sin 3x)/(tan 5x), (c) lim x->inf (3x2+5x2+1)/(3x3-2x)

Substituting $x=2$ into the expression $\frac{x^3 - 8}{x^2 - 4}$ gives:

This is an indeterminate form, so we need to simplify the expression.

The numerator is a difference of cubes: $x^3 - 8 = (x-2)(x^2 + 2x + 4)$.
The denominator is a difference of squares: $x^2 - 4 = (x-2)(x+2)$.

Since $x \to 2$, $x \neq 2$, so $x-2 \neq 0$. We can cancel the common factor $(x-2)$:

The value of the limit is 3.