Solve the following limit problems: Problem 1: lim x→0 (sqrt(x2+9)-3)/x2. Problem 2: lim x→1 (e^(x-1)-x)/(x-1)2.

$$\frac{\sqrt{0^2+9}-3}{0^2} = \frac{\sqrt{9}-3}{0} = \frac{3-3}{0} = \frac{0}{0}$$

This is an indeterminate form.

$$\lim_{x \to 0} \frac{\sqrt{x^2+9}-3}{x^2} \times \frac{\sqrt{x^2+9}+3}{\sqrt{x^2+9}+3}$$ $$= \lim_{x \to 0} \frac{(\sqrt{x^2+9})^2 - 3^2}{x^2(\sqrt{x^2+9}+3)}$$ $$= \lim_{x \to 0} \frac{x^2+9-9}{x^2(\sqrt{x^2+9}+3)}$$ $$= \lim_{x \to 0} \frac{x^2}{x^2(\sqrt{x^2+9}+3)}$$

$$= \frac{1}{\sqrt{0^2+9}+3} = \frac{1}{\sqrt{9}+3} = \frac{1}{3+3} = \frac{1}{6}$$

The limit is $\boxed{\frac{1}{6}}$.

:

$$\lim_{x \to 1} \frac{e^{x-1}-x}{(x-1)^2}$$

$$\frac{e^{1-1}-1}{(1-1)^2} = \frac{e^0-1}{0^2} = \frac{1-1}{0} = \frac{0}{0}$$

This is an indeterminate form, so we apply L'Hôpital's Rule.

Let $f(x) = e^{x-1}-x$ and $g(x) = (x-1)^2$.
$f'(x) = \frac{d}{dx}(e^{x-1}-x) = e^{x-1}-1$
$g'(x) = \frac{d}{dx}((x-1)^2) = 2(x-1)$

$$\lim_{x \to 1} \frac{e^{x-1}-1}{2(x-1)}$$

$$\frac{e^{1-1}-1}{2(1-1)} = \frac{e^0-1}{0} = \frac{1-1}{0} = \frac{0}{0}$$

This is still an indeterminate form, so we apply L'Hôpital's Rule again.

Let $f_2(x) = e^{x-1}-1$ and $g_2(x) = 2(x-1)$.
$f_2'(x) = \frac{d}{dx}(e^{x-1}-1) = e^{x-1}$
$g_2'(x) = \frac{d}{dx}(2(x-1)) = 2$

$$\lim_{x \to 1} \frac{e^{x-1}}{2}$$

$$= \frac{e^{1-1}}{2} = \frac{e^0}{2} = \frac{1}{2}$$

The limit is $\boxed{\frac{1}{2}}$.

:

$$\lim_{x \to 4} \frac{x-4}{\sqrt{x}-2}$$

$$\frac{4-4}{\sqrt{4}-2} = \frac{0}{2-2} = \frac{0}{0}$$

This is an indeterminate form.

We can write $x-4$ as $(\sqrt{x})^2 - 2^2$.

$$x-4 = (\sqrt{x}-2)(\sqrt{x}+2)$$

Substitute this back into the limit expression:

$$\lim_{x \to 4} \frac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}-2}$$

$$= \sqrt{4}+2 = 2+2 = 4$$

The limit is $\boxed{4}$.