a) Prove that lim(x->0) (sqrt(1+x)-1)/x = 1/2. b) Determine lim(x->0) (e^x + e^-x - 2)/x2.

*a) Prove that $\lim_{x \to 0} \frac{\sqrt{1+x}-1}{x} = \frac{1}{2}$.

Step 1: Check the form of the limit. Substitute $x=0$ into the expression: $\frac{\sqrt{1+0}-1}{0} = \frac{\sqrt{1}-1}{0} = \frac{1-1}{0} = \frac{0}{0}$ This is an indeterminate form, so we can use algebraic manipulation or L'Hôpital's Rule. We will use algebraic manipulation by multiplying by the conjugate.

Step 2: Multiply the numerator and denominator by the conjugate of the numerator. The conjugate of $\sqrt{1+x}-1$ is $\sqrt{1+x}+1$. $\lim_{x \to 0} \frac{\sqrt{1+x}-1}{x} = \lim_{x \to 0} \frac{\sqrt{1+x}-1}{x} \times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$

Step 3: Simplify the expression. Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$: $= \lim_{x \to 0} \frac{(\sqrt{1+x})^2 - 1^2}{x(\sqrt{1+x}+1)}$ $= \lim_{x \to 0} \frac{(1+x) - 1}{x(\sqrt{1+x}+1)}$ $= \lim_{x \to 0} \frac{x}{x(\sqrt{1+x}+1)}$

Step 4: Cancel the common factor $x$ (since $x \to 0$, $x \neq 0$). $= \lim_{x \to 0} \frac{1}{\sqrt{1+x}+1}$

Step 5: Substitute $x=0$ into the simplified expression. $= \frac{1}{\sqrt{1+0}+1}$ $= \frac{1}{\sqrt{1}+1}$ $= \frac{1}{1+1}$ $= \frac{1}{2}$ Thus, \lim_{x \to 0} \frac{\sqrt{1+x}-1}{x} = \frac{1{2}}.

*b) Determine $\lim_{x \to 0} \frac{e^x + e^{-x} - 2}{x^2}$.

Step 1: Check the form of the limit. Substitute $x=0$ into the expression: $\frac{e^0 + e^{-0} - 2}{0^2} = \frac{1 + 1 - 2}{0} = \frac{0}{0}$ This is an indeterminate form, so we can apply L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule for the first time. Let $f(x) = e^x + e^{-x} - 2$ and $g(x) = x^2$. Then $f'(x) = \frac{d}{dx}(e^x + e^{-x} - 2) = e^x - e^{-x}$. And $g'(x) = \frac{d}{dx}(x^2) = 2x$. $\lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{e^x - e^{-x}}{2x}$

Step 3: Check the form of the new limit. Substitute $x=0$ into the new expression: $\frac{e^0 - e^{-0}}{2(0)} = \frac{1 - 1}{0} = \frac{0}{0}$ This is still an indeterminate form, so we apply L'Hôpital's Rule again.

Step 4: Apply L'Hôpital's Rule for the second time. Let $f''(x) = \frac{d}{dx}(e^x - e^{-x}) = e^x - (-e^{-x}) = e^x + e^{-x}$. And $g''(x) = \frac{d}{dx}(2x) = 2$. $\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \lim_{x \to 0} \frac{e^x + e^{-x}}{2}$

Step 5: Substitute $x=0$ into the simplified expression. $= \frac{e^0 + e^{-0}}{2}$ $= \frac{1 + 1}{2}$ $= \frac{2}{2}$ $= 1$ The value of the limit is $\boxed{1}$.