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

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### 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 \(x \to 0, \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} \to 1\):
\[
\text{Result: } \frac{2}{4} = \frac{1}{2}.
\]

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### 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 \(x - 3\):
\[
\text{Result: } \frac{\sqrt{3 + 6} + 3}{\sqrt{3 + 1} + 2} = \frac{3 + 3}{2 + 2} = \frac{6}{4} = \frac{3}{2}.
\]

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### 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:
\[
\frac{\frac{2x^3}{6}}{\frac{x^3}{6}} = 2.
\]
\[
\text{Result: } 2.
\]

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### 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 \(x \to 0\):
\[
\text{Result: } 1.
\]

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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!

Calculate the following limits: 1) lim(x→0) [(1 - cos x)/x^2], 2) lim(x→3) [(√(x + 6) - 3)/(√(x + 1) - 2)], 3) lim(x→0) [(e^x - e^(-x) - 2x)/(x - sin x)], 4) lim(x→0) [(5x - tan(4x))/x]

Detailed solutions to advanced calculus problems on limits, including trigonometric and exponential forms. Explore concepts such as Taylor expansions, trigonometric identities, and L'Hôpital's Rule for indeterminate forms. Perfect for students studying Calculus I.

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