Here is the breakdown of solving the given limits step by step:

1. $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$

This is a standard limit problem. Using the trigonometric identity: $1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)$ Substituting and simplifying: $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x^2}$ $= \lim_{x \to 0} \frac{2 \left(\frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2}{4}$ As $x \to 0, \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} \to 1$: $\text{Result: } \frac{2}{4} = \frac{1}{2}.$

2. $\lim_{x \to 3} \frac{\sqrt{x + 6} - 3}{\sqrt{x + 1} - 2}$

To simplify, rationalize the numerator and denominator: $\lim_{x \to 3} \frac{\sqrt{x + 6} - 3}{\sqrt{x + 1} - 2} \cdot \frac{\sqrt{x + 6} + 3}{\sqrt{x + 1} + 2}$ This gives: $\lim_{x \to 3} \frac{(x + 6) - 9}{(x + 1) - 4} \cdot \frac{\sqrt{x + 6} + 3}{\sqrt{x + 1} + 2}$ Simplify: $\lim_{x \to 3} \frac{x - 3}{x - 3} \cdot \frac{\sqrt{x + 6} + 3}{\sqrt{x + 1} + 2}$ Cancel $x - 3$: $\text{Result: } \frac{\sqrt{3 + 6} + 3}{\sqrt{3 + 1} + 2} = \frac{3 + 3}{2 + 2} = \frac{6}{4} = \frac{3}{2}.$

3. $\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x}$

Using Taylor expansions for small $x$: $e^x = 1 + x + \frac{x^2}{2} + \dots, \quad e^{-x} = 1 - x + \frac{x^2}{2} - \dots$ $\sin x = x - \frac{x^3}{6} + \dots$ Substitute these expansions: $e^x - e^{-x} = 2x + \frac{2x^3}{6}, \quad x - \sin x = \frac{x^3}{6}.$ The numerator becomes: $(e^x - e^{-x} - 2x) = \frac{2x^3}{6}, \quad x - \sin x = \frac{x^3}{6}.$ Simplify: $\frac{\frac{2x^3}{6}}{\frac{x^3}{6}} = 2.$ $\text{Result: } 2.$

4. $\lim_{x \to 0} \frac{5x - \tan 4x}{x}$

Using the Taylor expansion for $\tan 4x$ for small $x$: $\tan 4x = 4x + \frac{(4x)^3}{3} + \dots$ The numerator becomes: $5x - \tan 4x = 5x - (4x + \frac{64x^3}{3}) = x - \frac{64x^3}{3}.$ Divide by $x$: $\frac{5x - \tan 4x}{x} = 1 - \frac{64x^2}{3}.$ As $x \to 0$: $\text{Result: } 1.$

Do you need further steps or clarifications?

Here are some follow-up questions for practice:

1. Can you find the derivative of $f(x) = \frac{1 - \cos x}{x^2}$ at $x = 0$?
2. How does rationalizing help simplify limits involving square roots?
3. Can you generalize the Taylor series for $\tan(kx)$ to approximate similar limits?
4. What happens if the numerator and denominator of a limit both tend to infinity? Which rule applies?
5. How do expansions help to find limits in indeterminate forms like $\frac{0}{0}$?

Tip: Always test for indeterminate forms ($\frac{0}{0}$, $\infty/\infty$) before applying rules like L'Hôpital's Rule!