Let f: (0, infinity) R be a twice differentiable function such that f(2)=18, f'(2)=0 and f''(2)=4. Find the limit _x 2 (f(x) - 18 - (x-2)2)/(f(x) - 18 - 2(x-2)2).

Here are the solutions to the questions you've provided.

Question 1:
Let $f: (0, \infty) \to \mathbb{R}$ be a twice differentiable function such that $f(2)=18$, $f'(2)=0$ and $f''(2)=4$. We need to find the limit:

$$\lim_{x \to 2} \frac{f(x) - 18 - (x-2)^2}{f(x) - 18 - 2(x-2)^2}$$

Step 1: Evaluate the numerator and denominator at $x=2$.
Numerator at $x=2$: $f(2) - 18 - (2-2)^2 =