Evaluate the limit: lim x->0 (sqrt(x2+9)-3)/x2 and lim x->1 (e^(x-1)-1)/(x-1)2

Substituting $x=0$ into the expression gives:

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

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

Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x^2+9}+3$:

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

Cancel out the $x^2$ term from the numerator and denominator (since $x \to 0$, $x \neq 0$):

$$= \lim_{x \to 0} \frac{1}{\sqrt{x^2+9}+3}$$

Substitute $x=0$ into the simplified expression:

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

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

Substituting $x=1$ into the expression gives:

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

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

Take the derivative of the numerator and the denominator:
Numerator: $\frac{d}{dx}(e^{x-1}-x) = e^{x-1} \cdot 1 - 1 = e^{x-1}-1$
Denominator: $\frac{d}{dx}((x-1)^2) = 2(x-1) \cdot 1 = 2(x-1)$
The limit becomes:

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

Substituting $x=1$ into the new expression gives:

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

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

Take the derivative of the new numerator and denominator:
Numerator: $\frac{d}{dx}(e^{x-1}-1) = e^{x-1} \cdot 1 - 0 = e^{x-1}$
Denominator: $\frac{d}{dx}(2(x-1)) = 2 \cdot 1 = 2$
The limit becomes:

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

Substitute $x=1$ into the expression:

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

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

Substituting $x=4$ into the expression gives:

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

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

Recognize that the numerator $x-4$ can be written as a difference of squares, $(\sqrt{x})^2 - 2^2$:

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

Substitute this back into the limit expression:

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

Cancel out the common term $(\sqrt{x}-2)$ from the numerator and denominator (since $x \to 4$, $\sqrt{x} \neq 2$):

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

Substitute $x=4$ into the simplified expression:

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

The limit is $\boxed{4}$.