Let's solve the limits step by step:

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

Solution: Using the Taylor expansion for $\cos x$ around $x = 0$: $\cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots$ So, $1 - \cos x \approx \frac{x^2}{2} - \frac{x^4}{24} + \cdots$

Substituting into the limit: $\frac{1 - \cos x}{x^2} \approx \frac{\frac{x^2}{2} - \frac{x^4}{24} + \cdots}{x^2}$ $= \frac{1}{2} - \frac{x^2}{24} + \cdots$

As $x \to 0$, higher-order terms vanish, so: $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}.$

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

Solution:

First, expand the numerator and denominator using Taylor expansions:

• $e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$,
• $e^{-x} \approx 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots$,
• $\sin x \approx x - \frac{x^3}{6} + \cdots$.

Numerator: $e^x - e^{-x} - 2x \approx \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots\right) - \left(1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots\right) - 2x$ $= 2x + \frac{x^3}{6} - \left(2x\right) = \frac{x^3}{6}.$

Denominator: $x - \sin x \approx x - \left(x - \frac{x^3}{6} + \cdots\right) = \frac{x^3}{6}.$

Thus, the limit becomes: $\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x} = \lim_{x \to 0} \frac{\frac{x^3}{6}}{\frac{x^3}{6}} = 1.$

Final Answers:

1. $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$,
2. $\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x} = 1$.

Would you like a more detailed explanation for any of these steps or additional assistance?

Related Questions:

1. How can Taylor series expansions be used in solving limits involving trigonometric functions?
2. What is the derivation of Taylor expansions for $e^x$ and $\cos x$?
3. Why does dividing higher-order terms by lower-order terms result in their vanishing in limits?
4. How would L'Hôpital's Rule apply to these problems, if at all?
5. What are some common limits involving $\sin x$, $\cos x$, and $e^x$ for small $x$?

Tip:

When solving limits with indeterminate forms, Taylor expansions often simplify the problem significantly!