Evaluate the following limits: 1. lim (x3 - 27)/(x2 - 9) as x approaches 3. 2. lim (sqrt(x+2) - 3)/(x-7) as x approaches 7.

$$
\frac{3^3 - 27}{3^2 - 9} = \frac{27 - 27}{9 - 9} = \frac{0}{0}
$$
This is an indeterminate form, so we need to simplify the expression.

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

$$
\frac{3^2 + 3(3) + 9}{3+3} = \frac{9 + 9 + 9}{6} = \frac{27}{6}
$$

$$
\frac{27}{6} = \frac{9}{2}
$$
The limit is $\boxed{\frac{9}{2}}$.

$$
\frac{\sqrt{7+2} - 3}{7-7} = \frac{\sqrt{9} - 3}{0} = \frac{3-3}{0} = \frac{0}{0}
$$
This is an indeterminate form, so we need to simplify the expression.

$$
\lim_{x \to 7} \frac{\sqrt{x+2} - 3}{x-7} \times \frac{\sqrt{x+2} + 3}{\sqrt{x+2} + 3}
$$

$$
(\sqrt{x+2})^2 - 3^2 = (x+2) - 9 = x-7
$$

$$
\lim_{x \to 7} \frac{x-7}{(x-7)(\sqrt{x+2} + 3)}
$$

$$
\lim_{x \to 7} \frac{1}{\sqrt{x+2} + 3}
$$

$$
\frac{1}{\sqrt{7+2} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3+3} = \frac{1}{6}
$$
The limit is $\boxed{\frac{1}{6}}$.

$$
\lim_{x \to \infty} \frac{4 - \frac{5}{x} + \frac{1}{x^2}}{\sqrt{9 + \frac{2}{x^2} - \frac{6}{x^4}}}
$$

$$
\frac{4}{3}
$$
The limit is $\boxed{\frac{4}{3}}$.

$$
\lim_{x \to 0} \frac{\sin(5x)}{\frac{\sin(3x)}{\cos(3x)}} = \lim_{x \to 0} \frac{\sin(5x) \cos(3x)}{\sin(3x)}
$$

$$
\lim_{x \to 0} \left( \frac{\sin(5x)}{5x} \cdot \frac{3x}{\sin(3x)} \cdot \frac{5x}{3x} \cdot \cos(3x) \right)
$$

$$
\lim_{x \to 0} \left( \frac{\sin(5x)}{5x} \right) \cdot \lim_{x \to 0} \left( \frac{3x}{\sin(3x)} \right) \cdot \lim_{x \to 0} \left( \frac{5}{3} \right) \cdot \lim_{x \to 0} \left( \cos(3x) \right)
$$

$$
1 \cdot 1 \cdot \frac{5}{3} \cdot \cos(0) = 1 \cdot 1 \cdot \frac{5}{3} \cdot 1 = \frac{5}{3}
$$
The limit is $\boxed{\frac{5}{3}}$.

*   First :
    $$
    \lim_{x \to 0} \frac{1 - \cos x}{x^2}
    $$

    $$
    \lim_{x \to 0} 2 \cdot \frac{\sin\left(\frac{x}{2}\right)}{x} \cdot \frac{\sin\left(\frac{x}{2}\right)}{x}
    $$

    $$
    \lim_{x \to 0} 2 \cdot \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2} \cdot 2} \cdot \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2} \cdot 2}
    $$

    $$
    \lim_{x \to 0} 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \left( \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} \right) \cdot \left( \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} \right) = 2 \cdot \frac{1}{4} \cdot 1 \cdot 1 = \frac{1}{2}
    $$
    Alternatively, using L'Hôpital's Rule (since substituting $x=0$ gives $\frac{0}{0}$):

    $$
    \lim_{x \to 0} \frac{\frac{d}{dx}(1 - \cos x)}{\frac{d}{dx}(x^2)} = \lim_{x \to 0} \frac{\sin x}{2x}
    $$

    $$
    \lim_{x \to 0} \frac{1}{2} \cdot \frac{\sin x}{x}
    $$

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

*   Second :
    $$
    \lim_{x \to 0} \frac{1 - \cos x}{x}
    $$

    $$
    \frac{1 - \cos 0}{0} = \frac{1 - 1}{0} = \frac{0}{0}
    $$
    This is an indeterminate form, so we can use L'Hôpital's Rule.

    $$
    \lim_{x \to 0} \frac{\frac{d}{dx}(1 - \cos x)}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{\sin x}{1}
    $$

    $$
    \sin(0) = 0
    $$
    The limit is $\boxed{0}$.