Evaluate lim_(x->0) ((x+b)-b)/x

First, simplify the numerator of the expression.

$$\lim_{x \to 0} \frac{(x+b)-b}{x} = \lim_{x \to 0} \frac{x}{x}$$

Since $x \to 0$, $x$ is not exactly zero, so we can cancel $x$ from the numerator and denominator.

$$\lim_{x \to 0} 1 = 1$$

Answer for 16: $\boxed{1}$

If we substitute $x=0$ directly, we get $\frac{0}{0}$, which is an indeterminate form. We need to simplify the expression.
Expand the numerator:

$$(x+3)^2 - 9 = (x^2 + 6x + 9) - 9 = x^2 + 6x$$

Now substitute this back into the limit expression:

$$\lim_{x \to 0} \frac{x^2 + 6x}{x}$$

Factor out $x$ from the numerator:

$$\lim_{x \to 0} \frac{x(x+6)}{x}$$

Since $x \to 0$, $x$ is not exactly zero, so we can cancel $x$:

$$\lim_{x \to 0} (x+6)$$

Now substitute $x=0$:

$$0+6 = 6$$

Answer for 17: $\boxed{6}$

If we substitute $h=0$ directly, we get $\frac{0}{0}$, which is an indeterminate form. We need to simplify the expression.
Expand the numerator $(x+h)^3$:

$$(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$$

Now substitute this back into the numerator of the limit expression:

$$(x^3 + 3x^2h + 3xh^2 + h^3) - x^3 = 3x^2h + 3xh^2 + h^3$$

Substitute this simplified numerator back into the limit:

$$\lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$$

Factor out $h$ from the numerator:

$$\lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h}$$

Since $h \to 0$, $h$ is not exactly zero, so we can cancel $h$:

$$\lim_{h \to 0} (3x^2 + 3xh + h^2)$$

Now substitute $h=0$:

$$3x^2 + 3x(0) + (0)^2 = 3x^2 + 0 + 0 = 3x^2$$

Answer for 18: $\boxed{3x^2}$